Exercise 3.3 Q1. Prove that: sin 2 π 6 + cos 2 π 3 − tan 2 π 4 = − 1 2 \sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}=-\frac12 sin 2 6 π + cos 2 3 π − tan 2 4 π = − 2 1
Sol. We know that sin π 6 = 1 2 \sin\frac{\pi}{6}=\frac12 sin 6 π = 2 1 , cos π 3 = 1 2 \cos\frac{\pi}{3}=\frac12 cos 3 π = 2 1 and tan π 4 = 1 \tan\frac{\pi}{4}=1 tan 4 π = 1 .
LHS = sin 2 π 6 + cos 2 π 3 − tan 2 π 4
\text{LHS}=\sin^2\frac{\pi}{6}+\cos^2\frac{\pi}{3}-\tan^2\frac{\pi}{4}
LHS = sin 2 6 π + cos 2 3 π − tan 2 4 π Substituting the values,
= ( 1 2 ) 2 + ( 1 2 ) 2 − ( 1 ) 2 \displaystyle =\left(\frac12\right)^2+\left(\frac12\right)^2-(1)^2 = ( 2 1 ) 2 + ( 2 1 ) 2 − ( 1 ) 2 = 1 4 + 1 4 − 1 \displaystyle =\frac14+\frac14-1 = 4 1 + 4 1 − 1 = − 1 2 = RHS \displaystyle =-\frac12=\text{RHS} = − 2 1 = RHS Hence proved.
Q2. Prove that: 2 sin 2 π 6 + cosec 2 7 π 6 cos 2 π 3 = 3 2 2\sin^2\frac{\pi}{6}+\text{cosec}^2\frac{7\pi}{6}\cos^2\frac{\pi}{3}=\frac32 2 sin 2 6 π + cosec 2 6 7 π cos 2 3 π = 2 3
Sol. Using cosec ( π + x ) = − cosec x \text{cosec}(\pi+x)=-\text{cosec}\,x cosec ( π + x ) = − cosec x ,
cosec 7 π 6 = cosec ( π + π 6 ) = − cosec π 6 = − 2
\text{cosec}\frac{7\pi}{6}=\text{cosec}\left(\pi+\frac{\pi}{6}\right)=-\text{cosec}\frac{\pi}{6}=-2
cosec 6 7 π = cosec ( π + 6 π ) = − cosec 6 π = − 2 Also, sin π 6 = 1 2 \sin\frac{\pi}{6}=\frac12 sin 6 π = 2 1 and cos π 3 = 1 2 \cos\frac{\pi}{3}=\frac12 cos 3 π = 2 1 .
LHS = 2 sin 2 π 6 + cosec 2 7 π 6 cos 2 π 3
\text{LHS}=2\sin^2\frac{\pi}{6}+\text{cosec}^2\frac{7\pi}{6}\cos^2\frac{\pi}{3}
LHS = 2 sin 2 6 π + cosec 2 6 7 π cos 2 3 π Substituting the values,
= 2 ( 1 2 ) 2 + ( − 2 ) 2 ( 1 2 ) 2 \displaystyle =2\left(\frac12\right)^2+(-2)^2\left(\frac12\right)^2 = 2 ( 2 1 ) 2 + ( − 2 ) 2 ( 2 1 ) 2 = 2 × 1 4 + 4 × 1 4 \displaystyle =2\times\frac14+4\times\frac14 = 2 × 4 1 + 4 × 4 1 = 1 2 + 1 \displaystyle =\frac12+1 = 2 1 + 1 = 3 2 = RHS \displaystyle =\frac32=\text{RHS} = 2 3 = RHS Hence proved.
Q3. Prove that: cot 2 π 6 + cosec 5 π 6 + 3 tan 2 π 6 = 6 \cot^2\frac{\pi}{6}+\text{cosec}\frac{5\pi}{6}+3\tan^2\frac{\pi}{6}=6 cot 2 6 π + cosec 6 5 π + 3 tan 2 6 π = 6
Sol. Using cosec ( π − x ) = cosec x \text{cosec}(\pi-x)=\text{cosec}\,x cosec ( π − x ) = cosec x ,
cosec 5 π 6 = cosec ( π − π 6 ) = cosec π 6 = 2
\text{cosec}\frac{5\pi}{6}=\text{cosec}\left(\pi-\frac{\pi}{6}\right)=\text{cosec}\frac{\pi}{6}=2
cosec 6 5 π = cosec ( π − 6 π ) = cosec 6 π = 2 Also, cot π 6 = 3 \cot\frac{\pi}{6}=\sqrt3 cot 6 π = 3 and tan π 6 = 1 3 \tan\frac{\pi}{6}=\frac{1}{\sqrt3} tan 6 π = 3 1 .
LHS = cot 2 π 6 + cosec 5 π 6 + 3 tan 2 π 6
\text{LHS}=\cot^2\frac{\pi}{6}+\text{cosec}\frac{5\pi}{6}+3\tan^2\frac{\pi}{6}
LHS = cot 2 6 π + cosec 6 5 π + 3 tan 2 6 π Substituting the values,
= ( 3 ) 2 + 2 + 3 ( 1 3 ) 2 \displaystyle =(\sqrt3)^2+2+3\left(\frac{1}{\sqrt3}\right)^2 = ( 3 ) 2 + 2 + 3 ( 3 1 ) 2 = 3 + 2 + 3 × 1 3 \displaystyle =3+2+3\times\frac13 = 3 + 2 + 3 × 3 1 = 3 + 2 + 1 \displaystyle =3+2+1 = 3 + 2 + 1 = 6 = RHS \displaystyle =6=\text{RHS} = 6 = RHS Hence proved.
Q4. Prove that: 2 sin 2 3 π 4 + 2 cos 2 π 4 + 2 sec 2 π 3 = 10 2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3}=10 2 sin 2 4 3 π + 2 cos 2 4 π + 2 sec 2 3 π = 10
Sol. Using sin ( π − x ) = sin x \sin(\pi-x)=\sin x sin ( π − x ) = sin x ,
sin 3 π 4 = sin ( π − π 4 ) = sin π 4 = 1 2
\sin\frac{3\pi}{4}=\sin\left(\pi-\frac{\pi}{4}\right)=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}
sin 4 3 π = sin ( π − 4 π ) = sin 4 π = 2 1 Also, cos π 4 = 1 2 \cos\frac{\pi}{4}=\frac{1}{\sqrt2} cos 4 π = 2 1 and sec π 3 = 2 \sec\frac{\pi}{3}=2 sec 3 π = 2 .
LHS = 2 sin 2 3 π 4 + 2 cos 2 π 4 + 2 sec 2 π 3
\text{LHS}=2\sin^2\frac{3\pi}{4}+2\cos^2\frac{\pi}{4}+2\sec^2\frac{\pi}{3}
LHS = 2 sin 2 4 3 π + 2 cos 2 4 π + 2 sec 2 3 π Substituting the values,
= 2 ( 1 2 ) 2 + 2 ( 1 2 ) 2 + 2 ( 2 ) 2 \displaystyle =2\left(\frac{1}{\sqrt2}\right)^2+2\left(\frac{1}{\sqrt2}\right)^2+2(2)^2 = 2 ( 2 1 ) 2 + 2 ( 2 1 ) 2 + 2 ( 2 ) 2 = 1 + 1 + 8 \displaystyle =1+1+8 = 1 + 1 + 8 = 10 = RHS \displaystyle =10=\text{RHS} = 10 = RHS Hence proved.
(i) sin 75 ∘ \sin75^\circ sin 7 5 ∘
(ii) tan 15 ∘ \tan15^\circ tan 1 5 ∘
(i) Sol. Writing 75 ∘ = 45 ∘ + 30 ∘ 75^\circ=45^\circ+30^\circ 7 5 ∘ = 4 5 ∘ + 3 0 ∘ ,
sin 75 ∘ = sin ( 45 ∘ + 30 ∘ )
\begin{array}{l}
\displaystyle \sin75^\circ=\sin(45^\circ+30^\circ)
\end{array}
sin 7 5 ∘ = sin ( 4 5 ∘ + 3 0 ∘ ) Using the identity
sin ( x + y ) = sin x cos y + cos x sin y \textcolor{#1B5E20}{
\begin{array}{l}
\displaystyle \sin(x+y)=\sin x\cos y+\cos x\sin y
\end{array}
} s i n ( x + y ) = s i n x c o s y + c o s x s i n y = sin 45 ∘ cos 30 ∘ + cos 45 ∘ sin 30 ∘
\begin{array}{l}
\hspace{0.8em} \displaystyle =\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ
\end{array}
= sin 4 5 ∘ cos 3 0 ∘ + cos 4 5 ∘ sin 3 0 ∘ Substituting the values,
= 1 2 × 3 2 + 1 2 × 1 2 = 3 + 1 2 2
\begin{array}{l}
\hspace{2.6em} \displaystyle =\frac{1}{\sqrt2}\times\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\times\frac12 \\[0.8em]
\hspace{2.6em} \displaystyle =\frac{\sqrt3+1}{2\sqrt2}
\end{array}
= 2 1 × 2 3 + 2 1 × 2 1 = 2 2 3 + 1 (ii) Sol. Writing 15 ∘ = 45 ∘ − 30 ∘ 15^\circ=45^\circ-30^\circ 1 5 ∘ = 4 5 ∘ − 3 0 ∘ ,
tan 15 ∘ = tan ( 45 ∘ − 30 ∘ )
\begin{array}{l}
\displaystyle \tan15^\circ=\tan(45^\circ-30^\circ)
\end{array}
tan 1 5 ∘ = tan ( 4 5 ∘ − 3 0 ∘ ) Using the identity
tan ( x − y ) = tan x − tan y 1 + tan x tan y \textcolor{#1B5E20}{
\begin{array}{l}
\displaystyle \tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}
\end{array}
} t a n ( x − y ) = 1 + t a n x t a n y t a n x − t a n y = tan 45 ∘ − tan 30 ∘ 1 + tan 45 ∘ tan 30 ∘
\begin{array}{l}
\hspace{0.8em} \displaystyle =\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}
\end{array}
= 1 + tan 4 5 ∘ tan 3 0 ∘ tan 4 5 ∘ − tan 3 0 ∘ Substituting the values,
= 1 − 1 3 1 + 1 × 1 3 = 3 − 1 3 + 1
\begin{array}{l}
\hspace{2.8em} \displaystyle =\frac{1-\dfrac{1}{\sqrt3}}{1+1\times\dfrac{1}{\sqrt3}} \\[0.8em]
\hspace{2.8em} \displaystyle =\frac{\sqrt3-1}{\sqrt3+1}
\end{array}
= 1 + 1 × 3 1 1 − 3 1 = 3 + 1 3 − 1 Rationalising the denominator,
= 3 − 1 3 + 1 × 3 − 1 3 − 1 = ( 3 − 1 ) 2 ( 3 ) 2 − 1 2 = 3 + 1 − 2 3 3 − 1 = 4 − 2 3 2 = 2 − 3
\begin{array}{l}
\hspace{2.8em} \displaystyle =\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1} \\[1em]
\hspace{2.8em} \displaystyle =\frac{(\sqrt3-1)^2}{(\sqrt3)^2-1^2} \\[1em]
\hspace{2.8em} \displaystyle =\frac{3+1-2\sqrt3}{3-1} \\[1em]
\hspace{2.8em} \displaystyle =\frac{4-2\sqrt3}{2}=2-\sqrt3
\end{array}
= 3 + 1 3 − 1 × 3 − 1 3 − 1 = ( 3 ) 2 − 1 2 ( 3 − 1 ) 2 = 3 − 1 3 + 1 − 2 3 = 2 4 − 2 3 = 2 − 3 cos ( π 4 − x ) cos ( π 4 − y ) − sin ( π 4 − x ) sin ( π 4 − y ) = sin ( x + y ) \textcolor{#C34E36}{
\begin{array}{l}
\cos\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-y\right) \\[0.6em]
\hspace{0.8em} -\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}-y\right)=\sin(x+y)
\end{array}
} c o s ( 4 π − x ) c o s ( 4 π − y ) − s i n ( 4 π − x ) s i n ( 4 π − y ) = s i n ( x + y ) Sol. We know that
cos A cos B − sin A sin B = cos ( A + B ) . . . ( i ) \textcolor{#1B5E20}{
\cos A\cos B-\sin A\sin B=\cos(A+B)\qquad ...(i)
} c o s A c o s B − s i n A s i n B = c o s ( A + B ) ... ( i ) Taking A = π 4 − x A=\frac{\pi}{4}-x A = 4 π − x and B = π 4 − y B=\frac{\pi}{4}-y B = 4 π − y ,
LHS = cos ( π 4 − x ) cos ( π 4 − y ) LHS = − sin ( π 4 − x ) sin ( π 4 − y )
\begin{array}{l}
\displaystyle \text{LHS}=\cos\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-y\right) \\[0.8em]
\displaystyle \phantom{\text{LHS}=}-\sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}-y\right)
\end{array}
LHS = cos ( 4 π − x ) cos ( 4 π − y ) LHS = − sin ( 4 π − x ) sin ( 4 π − y ) Using (i),
= cos [ ( π 4 − x ) + ( π 4 − y ) ]
\begin{array}{l}
\hspace{1.4em} \displaystyle =\cos\left[\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right]
\end{array}
= cos [ ( 4 π − x ) + ( 4 π − y ) ] = cos [ π 2 − ( x + y ) ]
\begin{array}{l}
\hspace{1.4em} \displaystyle =\cos\left[\frac{\pi}{2}-(x+y)\right]
\end{array}
= cos [ 2 π − ( x + y ) ] Using cos ( π 2 − θ ) = sin θ \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta cos ( 2 π − θ ) = sin θ ,
= sin ( x + y ) = RHS
=\sin(x+y)=\text{RHS}
= sin ( x + y ) = RHS Hence proved.
Q7. Prove the following: tan ( π 4 + x ) tan ( π 4 − x ) = ( 1 + tan x 1 − tan x ) 2 \frac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^2 t a n ( 4 π − x ) t a n ( 4 π + x ) = ( 1 − t a n x 1 + t a n x ) 2
Sol. We know that
tan ( A + B ) = tan A + tan B 1 − tan A tan B \textcolor{#1B5E20}{
\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}
} t a n ( A + B ) = 1 − t a n A t a n B t a n A + t a n B tan ( A − B ) = tan A − tan B 1 + tan A tan B \textcolor{#1B5E20}{
\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
} t a n ( A − B ) = 1 + t a n A t a n B t a n A − t a n B Taking A = π 4 A=\frac{\pi}{4} A = 4 π and B = x B=x B = x ,
tan ( π 4 + x ) = tan π 4 + tan x 1 − tan π 4 tan x = 1 + tan x 1 − tan x
\begin{array}{l}
\displaystyle \tan\left(\frac{\pi}{4}+x\right)=\frac{\tan\dfrac{\pi}{4}+\tan x}{1-\tan\dfrac{\pi}{4}\tan x} \\[1.6em]
\hspace{5.2em} \displaystyle =\frac{1+\tan x}{1-\tan x}
\end{array}
tan ( 4 π + x ) = 1 − tan 4 π tan x tan 4 π + tan x = 1 − tan x 1 + tan x Similarly,
tan ( π 4 − x ) = tan π 4 − tan x 1 + tan π 4 tan x = 1 − tan x 1 + tan x
\begin{array}{l}
\displaystyle \tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\dfrac{\pi}{4}-\tan x}{1+\tan\dfrac{\pi}{4}\tan x} \\[1.6em]
\hspace{5.2em} \displaystyle =\frac{1-\tan x}{1+\tan x}
\end{array}
tan ( 4 π − x ) = 1 + tan 4 π tan x tan 4 π − tan x = 1 + tan x 1 − tan x Substituting these expressions in the LHS,
LHS = tan ( π 4 + x ) tan ( π 4 − x ) = 1 + tan x 1 − tan x 1 − tan x 1 + tan x = 1 + tan x 1 − tan x × 1 + tan x 1 − tan x = ( 1 + tan x 1 − tan x ) 2 = RHS
\begin{array}{l}
\displaystyle \text{LHS}=\frac{\tan\left(\dfrac{\pi}{4}+x\right)}{\tan\left(\dfrac{\pi}{4}-x\right)} \\[1.8em]
\hspace{1.8em} \displaystyle =\frac{\dfrac{1+\tan x}{1-\tan x}}{\dfrac{1-\tan x}{1+\tan x}} \\[2em]
\hspace{1.8em} \displaystyle =\frac{1+\tan x}{1-\tan x}\times\frac{1+\tan x}{1-\tan x} \\[1em]
\hspace{1.8em} \displaystyle =\left(\frac{1+\tan x}{1-\tan x}\right)^2=\text{RHS}
\end{array}
LHS = tan ( 4 π − x ) tan ( 4 π + x ) = 1 + tan x 1 − tan x 1 − tan x 1 + tan x = 1 − tan x 1 + tan x × 1 − tan x 1 + tan x = ( 1 − tan x 1 + tan x ) 2 = RHS Hence proved.
Q8. Prove the following: cos ( π + x ) cos ( − x ) sin ( π − x ) cos ( π 2 + x ) = cot 2 x \frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\frac{\pi}{2}+x\right)}=\cot^2x s i n ( π − x ) c o s ( 2 π + x ) c o s ( π + x ) c o s ( − x ) = cot 2 x
Sol. We know that cos ( π + x ) = − cos x \cos(\pi+x)=-\cos x cos ( π + x ) = − cos x , cos ( − x ) = cos x \cos(-x)=\cos x cos ( − x ) = cos x , sin ( π − x ) = sin x \sin(\pi-x)=\sin x sin ( π − x ) = sin x and cos ( π 2 + x ) = − sin x \cos\left(\frac{\pi}{2}+x\right)=-\sin x cos ( 2 π + x ) = − sin x .
LHS = cos ( π + x ) cos ( − x ) sin ( π − x ) cos ( π 2 + x )
\text{LHS}=\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\dfrac{\pi}{2}+x\right)}
LHS = sin ( π − x ) cos ( 2 π + x ) cos ( π + x ) cos ( − x ) Substituting these,
= ( − cos x ) ( cos x ) ( sin x ) ( − sin x ) = cos 2 x sin 2 x = cot 2 x = RHS
\begin{array}{l}
\hspace{1.2em} \displaystyle =\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \\[0.8em]
\hspace{1.2em} \displaystyle =\frac{\cos^2x}{\sin^2x} \\[0.8em]
\hspace{1.2em} \displaystyle =\cot^2x=\text{RHS}
\end{array}
= ( sin x ) ( − sin x ) ( − cos x ) ( cos x ) = sin 2 x cos 2 x = cot 2 x = RHS Hence proved.
cos ( 3 π 2 + x ) cos ( 2 π + x ) [ cot ( 3 π 2 − x ) + cot ( 2 π + x ) ] = 1 \textcolor{#C34E36}{
\begin{array}{l}
\displaystyle \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x) \\[0.8em]
\hspace{1.8em} \displaystyle \left[\cot\left(\frac{3\pi}{2}-x\right)+\cot(2\pi+x)\right]=1
\end{array}
} c o s ( 2 3 π + x ) c o s ( 2 π + x ) [ c o t ( 2 3 π − x ) + c o t ( 2 π + x ) ] = 1 Sol. We know that cos ( 3 π 2 + x ) = sin x \cos\left(\frac{3\pi}{2}+x\right)=\sin x cos ( 2 3 π + x ) = sin x , cos ( 2 π + x ) = cos x \cos(2\pi+x)=\cos x cos ( 2 π + x ) = cos x , cot ( 3 π 2 − x ) = tan x \cot\left(\frac{3\pi}{2}-x\right)=\tan x cot ( 2 3 π − x ) = tan x and cot ( 2 π + x ) = cot x \cot(2\pi+x)=\cot x cot ( 2 π + x ) = cot x .
LHS = cos ( 3 π 2 + x ) cos ( 2 π + x ) [ cot ( 3 π 2 − x ) + cot ( 2 π + x ) ]
\text{LHS}=\cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x)\left[\cot\left(\frac{3\pi}{2}-x\right)+\cot(2\pi+x)\right]
LHS = cos ( 2 3 π + x ) cos ( 2 π + x ) [ cot ( 2 3 π − x ) + cot ( 2 π + x ) ] Substituting these,
= sin x cos x ( tan x + cot x ) \displaystyle =\sin x\cos x(\tan x+\cot x) = sin x cos x ( tan x + cot x ) = sin x cos x ( sin x cos x + cos x sin x ) \displaystyle =\sin x\cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right) = sin x cos x ( cos x sin x + sin x cos x ) = sin x cos x × sin 2 x + cos 2 x sin x cos x \displaystyle =\sin x\cos x\times\frac{\sin^2x+\cos^2x}{\sin x\cos x} = sin x cos x × sin x cos x sin 2 x + cos 2 x = sin 2 x + cos 2 x \displaystyle =\sin^2x+\cos^2x = sin 2 x + cos 2 x Using sin 2 x + cos 2 x = 1 \sin^2x+\cos^2x=1 sin 2 x + cos 2 x = 1 ,
= 1 = RHS
\begin{array}{l}
\hspace{1em} \displaystyle =1=\text{RHS}
\end{array}
= 1 = RHS Hence proved.
Q10. Prove the following:
sin ( n + 1 ) x sin ( n + 2 ) x + cos ( n + 1 ) x cos ( n + 2 ) x = cos x \textcolor{#C34E36}{
\begin{array}{l}
\displaystyle \sin(n+1)x\sin(n+2)x \\[0.4em]
\hspace{1.8em} \displaystyle +\cos(n+1)x\cos(n+2)x=\cos x
\end{array}
} s i n ( n + 1 ) x s i n ( n + 2 ) x + c o s ( n + 1 ) x c o s ( n + 2 ) x = c o s x Sol. We know that
sin A sin B + cos A cos B = cos ( A − B ) . . . ( i ) \textcolor{#1B5E20}{
\sin A\sin B+\cos A\cos B=\cos(A-B)\qquad ...(i)
} s i n A s i n B + c o s A c o s B = c o s ( A − B ) ... ( i ) Taking A = ( n + 2 ) x A=(n+2)x A = ( n + 2 ) x and B = ( n + 1 ) x B=(n+1)x B = ( n + 1 ) x ,
LHS = sin ( n + 1 ) x sin ( n + 2 ) x LHS = + cos ( n + 1 ) x cos ( n + 2 ) x
\begin{array}{l}
\displaystyle \text{LHS}=\sin(n+1)x\sin(n+2)x \\[0.4em]
\displaystyle \phantom{\text{LHS}=}+\cos(n+1)x\cos(n+2)x
\end{array}
LHS = sin ( n + 1 ) x sin ( n + 2 ) x LHS = + cos ( n + 1 ) x cos ( n + 2 ) x Using (i),
= cos [ ( n + 2 ) x − ( n + 1 ) x ]
\begin{array}{l}
\hspace{1.8em} \displaystyle =\cos[(n+2)x-(n+1)x]
\end{array}
= cos [( n + 2 ) x − ( n + 1 ) x ] = cos x = RHS
\begin{array}{l}
\hspace{1.8em} \displaystyle =\cos x=\text{RHS}
\end{array}
= cos x = RHS Hence proved.
Q11. Prove the following: cos ( 3 π 4 + x ) − cos ( 3 π 4 − x ) = − 2 sin x \cos\left(\frac{3\pi}{4}+x\right)-\cos\left(\frac{3\pi}{4}-x\right)=-\sqrt2\sin x cos ( 4 3 π + x ) − cos ( 4 3 π − x ) = − 2 sin x
Sol. We know that
cos A − cos B = − 2 sin A + B 2 sin A − B 2 . . . ( i ) \textcolor{#1B5E20}{
\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\qquad ...(i)
} c o s A − c o s B = − 2 s i n 2 A + B s i n 2 A − B ... ( i ) Taking A = 3 π 4 + x A=\frac{3\pi}{4}+x A = 4 3 π + x and B = 3 π 4 − x B=\frac{3\pi}{4}-x B = 4 3 π − x ,
LHS = cos ( 3 π 4 + x ) − cos ( 3 π 4 − x )
\begin{array}{l}
\displaystyle \text{LHS}=\cos\left(\frac{3\pi}{4}+x\right)-\cos\left(\frac{3\pi}{4}-x\right)
\end{array}
LHS = cos ( 4 3 π + x ) − cos ( 4 3 π − x ) Using (i),
= − 2 sin A + B 2 sin A − B 2
\begin{array}{l}
\hspace{1.8em} \displaystyle =-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\end{array}
= − 2 sin 2 A + B sin 2 A − B Now,
A + B = ( 3 π 4 + x ) + ( 3 π 4 − x ) = 3 π 2 A − B = ( 3 π 4 + x ) − ( 3 π 4 − x ) = 2 x
\begin{array}{l}
\displaystyle A+B=\left(\frac{3\pi}{4}+x\right)+\left(\frac{3\pi}{4}-x\right)=\frac{3\pi}{2} \\[1em]
\displaystyle A-B=\left(\frac{3\pi}{4}+x\right)-\left(\frac{3\pi}{4}-x\right)=2x
\end{array}
A + B = ( 4 3 π + x ) + ( 4 3 π − x ) = 2 3 π A − B = ( 4 3 π + x ) − ( 4 3 π − x ) = 2 x So, A + B 2 = 3 π 4 \frac{A+B}{2}=\frac{3\pi}{4} 2 A + B = 4 3 π and A − B 2 = x \frac{A-B}{2}=x 2 A − B = x . . . . ( i i ) \qquad ...(ii) ... ( ii )
Using (ii),
LHS = − 2 sin 3 π 4 sin x
\text{LHS}=-2\sin\frac{3\pi}{4}\sin x
LHS = − 2 sin 4 3 π sin x Using sin 3 π 4 = sin ( π − π 4 ) = sin π 4 = 1 2 \sin\frac{3\pi}{4}=\sin\left(\pi-\frac{\pi}{4}\right)=\sin\frac{\pi}{4}=\frac{1}{\sqrt2} sin 4 3 π = sin ( π − 4 π ) = sin 4 π = 2 1 ,
= − 2 × 1 2 sin x = − 2 sin x = RHS
\begin{array}{l}
\hspace{0.8em} \displaystyle =-2\times\frac{1}{\sqrt2}\sin x \\[0.8em]
\hspace{0.8em} \displaystyle =-\sqrt2\sin x=\text{RHS}
\end{array}
= − 2 × 2 1 sin x = − 2 sin x = RHS Hence proved.
Q12. Prove the following: sin 2 6 x − sin 2 4 x = sin 2 x sin 10 x \sin^26x-\sin^24x=\sin2x\sin10x sin 2 6 x − sin 2 4 x = sin 2 x sin 10 x
Sol.
LHS = sin 2 6 x − sin 2 4 x
\begin{array}{l}
\displaystyle \text{LHS}=\sin^26x-\sin^24x
\end{array}
LHS = sin 2 6 x − sin 2 4 x Using a 2 − b 2 = ( a + b ) ( a − b ) a^2-b^2=(a+b)(a-b) a 2 − b 2 = ( a + b ) ( a − b ) ,
= ( sin 6 x + sin 4 x ) ( sin 6 x − sin 4 x )
\begin{array}{l}
\hspace{1.8em} \displaystyle =(\sin6x+\sin4x)(\sin6x-\sin4x)
\end{array}
= ( sin 6 x + sin 4 x ) ( sin 6 x − sin 4 x ) Using sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B
and sin A − sin B = 2 cos A + B 2 sin A − B 2 \sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2} sin A − sin B = 2 cos 2 A + B sin 2 A − B ,
= ( 2 sin 6 x + 4 x 2 cos 6 x − 4 x 2 ) ( 2 cos 6 x + 4 x 2 sin 6 x − 4 x 2 ) \displaystyle =\left(2\sin\frac{6x+4x}{2}\cos\frac{6x-4x}{2}\right)\left(2\cos\frac{6x+4x}{2}\sin\frac{6x-4x}{2}\right) = ( 2 sin 2 6 x + 4 x cos 2 6 x − 4 x ) ( 2 cos 2 6 x + 4 x sin 2 6 x − 4 x ) = ( 2 sin 5 x cos x ) ( 2 cos 5 x sin x ) \displaystyle =(2\sin5x\cos x)(2\cos5x\sin x) = ( 2 sin 5 x cos x ) ( 2 cos 5 x sin x ) = ( 2 sin 5 x cos 5 x ) ( 2 sin x cos x ) \displaystyle =(2\sin5x\cos5x)(2\sin x\cos x) = ( 2 sin 5 x cos 5 x ) ( 2 sin x cos x ) Using 2 sin θ cos θ = sin 2 θ 2\sin\theta\cos\theta=\sin2\theta 2 sin θ cos θ = sin 2 θ ,
= sin 10 x sin 2 x = sin 2 x sin 10 x = RHS
\begin{array}{l}
\hspace{1.8em} \displaystyle =\sin10x\sin2x \\[0.4em]
\hspace{1.8em} \displaystyle =\sin2x\sin10x=\text{RHS}
\end{array}
= sin 10 x sin 2 x = sin 2 x sin 10 x = RHS Hence proved.
Q13. Prove the following: cos 2 2 x − cos 2 6 x = sin 4 x sin 8 x \cos^22x-\cos^26x=\sin4x\sin8x cos 2 2 x − cos 2 6 x = sin 4 x sin 8 x
Sol.
LHS = cos 2 2 x − cos 2 6 x
\begin{array}{l}
\displaystyle \text{LHS}=\cos^22x-\cos^26x
\end{array}
LHS = cos 2 2 x − cos 2 6 x Using a 2 − b 2 = ( a + b ) ( a − b ) a^2-b^2=(a+b)(a-b) a 2 − b 2 = ( a + b ) ( a − b ) ,
= ( cos 2 x + cos 6 x ) ( cos 2 x − cos 6 x )
\begin{array}{l}
\displaystyle =(\cos2x+\cos6x)(\cos2x-\cos6x)
\end{array}
= ( cos 2 x + cos 6 x ) ( cos 2 x − cos 6 x ) Using cos A + cos B = 2 cos A + B 2 cos A − B 2 \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2} cos A + cos B = 2 cos 2 A + B cos 2 A − B
and cos A − cos B = − 2 sin A + B 2 sin A − B 2 \cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2} cos A − cos B = − 2 sin 2 A + B sin 2 A − B ,
= [ 2 cos 2 x + 6 x 2 cos 2 x − 6 x 2 ] [ − 2 sin 2 x + 6 x 2 sin 2 x − 6 x 2 ] \displaystyle =\left[2\cos\frac{2x+6x}{2}\cos\frac{2x-6x}{2}\right]\left[-2\sin\frac{2x+6x}{2}\sin\frac{2x-6x}{2}\right] = [ 2 cos 2 2 x + 6 x cos 2 2 x − 6 x ] [ − 2 sin 2 2 x + 6 x sin 2 2 x − 6 x ] = [ 2 cos 4 x cos ( − 2 x ) ] [ − 2 sin 4 x sin ( − 2 x ) ] \displaystyle =[2\cos4x\cos(-2x)][-2\sin4x\sin(-2x)] = [ 2 cos 4 x cos ( − 2 x )] [ − 2 sin 4 x sin ( − 2 x )] Using cos ( − θ ) = cos θ \cos(-\theta)=\cos\theta cos ( − θ ) = cos θ and sin ( − θ ) = − sin θ \sin(-\theta)=-\sin\theta sin ( − θ ) = − sin θ ,
= ( 2 cos 4 x cos 2 x ) ( 2 sin 4 x sin 2 x ) = ( 2 sin 4 x cos 4 x ) ( 2 sin 2 x cos 2 x )
\begin{array}{l}
\displaystyle =(2\cos4x\cos2x)(2\sin4x\sin2x) \\[0.4em]
\displaystyle =(2\sin4x\cos4x)(2\sin2x\cos2x)
\end{array}
= ( 2 cos 4 x cos 2 x ) ( 2 sin 4 x sin 2 x ) = ( 2 sin 4 x cos 4 x ) ( 2 sin 2 x cos 2 x ) Using 2 sin θ cos θ = sin 2 θ 2\sin\theta\cos\theta=\sin2\theta 2 sin θ cos θ = sin 2 θ ,
= sin 8 x sin 4 x = sin 4 x sin 8 x = RHS
\begin{array}{l}
\displaystyle =\sin8x\sin4x \\[0.4em]
\displaystyle =\sin4x\sin8x=\text{RHS}
\end{array}
= sin 8 x sin 4 x = sin 4 x sin 8 x = RHS Hence proved.
Q14. Prove the following: sin 2 x + 2 sin 4 x + sin 6 x = 4 cos 2 x sin 4 x \sin2x+2\sin4x+\sin6x=4\cos^2x\sin4x sin 2 x + 2 sin 4 x + sin 6 x = 4 cos 2 x sin 4 x
Sol.
LHS = ( sin 6 x + sin 2 x ) + 2 sin 4 x
\text{LHS}=(\sin6x+\sin2x)+2\sin4x
LHS = ( sin 6 x + sin 2 x ) + 2 sin 4 x Using sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B ,
= 2 sin 6 x + 2 x 2 cos 6 x − 2 x 2 + 2 sin 4 x = 2 sin 4 x cos 2 x + 2 sin 4 x
\begin{array}{l}
\displaystyle =2\sin\frac{6x+2x}{2}\cos\frac{6x-2x}{2}+2\sin4x \\[0.8em]
\displaystyle =2\sin4x\cos2x+2\sin4x
\end{array}
= 2 sin 2 6 x + 2 x cos 2 6 x − 2 x + 2 sin 4 x = 2 sin 4 x cos 2 x + 2 sin 4 x = 2 sin 4 x ( cos 2 x + 1 )
=2\sin4x(\cos2x+1)
= 2 sin 4 x ( cos 2 x + 1 ) Using cos 2 x = 2 cos 2 x − 1 \cos2x=2\cos^2x-1 cos 2 x = 2 cos 2 x − 1 ,
= 2 sin 4 x ( 2 cos 2 x − 1 + 1 ) = 2 sin 4 x × 2 cos 2 x = 4 cos 2 x sin 4 x = RHS
\begin{array}{l}
\displaystyle =2\sin4x(2\cos^2x-1+1) \\[0.4em]
\displaystyle =2\sin4x\times2\cos^2x \\[0.4em]
\displaystyle =4\cos^2x\sin4x=\text{RHS}
\end{array}
= 2 sin 4 x ( 2 cos 2 x − 1 + 1 ) = 2 sin 4 x × 2 cos 2 x = 4 cos 2 x sin 4 x = RHS Hence proved.
Q15. Prove the following: cot 4 x ( sin 5 x + sin 3 x ) = cot x ( sin 5 x − sin 3 x ) \cot4x(\sin5x+\sin3x)=\cot x(\sin5x-\sin3x) cot 4 x ( sin 5 x + sin 3 x ) = cot x ( sin 5 x − sin 3 x )
Sol.
LHS = cot 4 x ( sin 5 x + sin 3 x )
\text{LHS}=\cot4x(\sin5x+\sin3x)
LHS = cot 4 x ( sin 5 x + sin 3 x ) Using sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B ,
= cot 4 x × 2 sin 5 x + 3 x 2 cos 5 x − 3 x 2 = cot 4 x ( 2 sin 4 x cos x ) = cos 4 x sin 4 x × 2 sin 4 x cos x = 2 cos 4 x cos x
\begin{array}{l}
\displaystyle =\cot4x\times2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2} \\[0.8em]
\displaystyle =\cot4x(2\sin4x\cos x) \\[0.4em]
\displaystyle =\frac{\cos4x}{\sin4x}\times2\sin4x\cos x \\[0.8em]
\displaystyle =2\cos4x\cos x
\end{array}
= cot 4 x × 2 sin 2 5 x + 3 x cos 2 5 x − 3 x = cot 4 x ( 2 sin 4 x cos x ) = sin 4 x cos 4 x × 2 sin 4 x cos x = 2 cos 4 x cos x RHS = cot x ( sin 5 x − sin 3 x )
\begin{array}{l}
\displaystyle \text{RHS}=\cot x(\sin5x-\sin3x)
\end{array}
RHS = cot x ( sin 5 x − sin 3 x ) Using sin A − sin B = 2 cos A + B 2 sin A − B 2 \sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2} sin A − sin B = 2 cos 2 A + B sin 2 A − B ,
= cos x sin x [ 2 cos 5 x + 3 x 2 sin 5 x − 3 x 2 ]
\begin{array}{l}
\hspace{0.8em} \displaystyle =\frac{\cos x}{\sin x}\left[2\cos\frac{5x+3x}{2}\sin\frac{5x-3x}{2}\right]
\end{array}
= sin x cos x [ 2 cos 2 5 x + 3 x sin 2 5 x − 3 x ] = cos x sin x [ 2 cos 4 x sin x ] = 2 cos 4 x cos x = LHS
\begin{array}{l}
\hspace{0.8em} \displaystyle =\frac{\cos x}{\sin x}\left[2\cos4x\sin x\right] \\[0.6em]
\hspace{0.8em} \displaystyle =2\cos4x\cos x=\text{LHS}
\end{array}
= sin x cos x [ 2 cos 4 x sin x ] = 2 cos 4 x cos x = LHS Hence proved.
Q16. Prove the following: cos 9 x − cos 5 x sin 17 x − sin 3 x = − sin 2 x cos 10 x \frac{\cos9x-\cos5x}{\sin17x-\sin3x}=-\frac{\sin2x}{\cos10x} s i n 17 x − s i n 3 x c o s 9 x − c o s 5 x = − c o s 10 x s i n 2 x
Sol. Using cos A − cos B = − 2 sin A + B 2 sin A − B 2 \cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2} cos A − cos B = − 2 sin 2 A + B sin 2 A − B
and sin A − sin B = 2 cos A + B 2 sin A − B 2 \sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2} sin A − sin B = 2 cos 2 A + B sin 2 A − B ,
LHS = cos 9 x − cos 5 x sin 17 x − sin 3 x = − 2 sin 9 x + 5 x 2 sin 9 x − 5 x 2 2 cos 17 x + 3 x 2 sin 17 x − 3 x 2 = − 2 sin 7 x sin 2 x 2 cos 10 x sin 7 x = − sin 2 x cos 10 x = RHS
\begin{array}{l}
\displaystyle \text{LHS}=\frac{\cos9x-\cos5x}{\sin17x-\sin3x} \\[1.2em]
\hspace{1.8em} \displaystyle =\frac{-2\sin\dfrac{9x+5x}{2}\sin\dfrac{9x-5x}{2}}{2\cos\dfrac{17x+3x}{2}\sin\dfrac{17x-3x}{2}} \\[2em]
\hspace{1.8em} \displaystyle =\frac{-2\sin7x\sin2x}{2\cos10x\sin7x} \\[1em]
\hspace{1.8em} \displaystyle =-\frac{\sin2x}{\cos10x}=\text{RHS}
\end{array}
LHS = sin 17 x − sin 3 x cos 9 x − cos 5 x = 2 cos 2 17 x + 3 x sin 2 17 x − 3 x − 2 sin 2 9 x + 5 x sin 2 9 x − 5 x = 2 cos 10 x sin 7 x − 2 sin 7 x sin 2 x = − cos 10 x sin 2 x = RHS Hence proved.
Q17. Prove the following: sin 5 x + sin 3 x cos 5 x + cos 3 x = tan 4 x \frac{\sin5x+\sin3x}{\cos5x+\cos3x}=\tan4x c o s 5 x + c o s 3 x s i n 5 x + s i n 3 x = tan 4 x
Sol. Using sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B
and cos A + cos B = 2 cos A + B 2 cos A − B 2 \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2} cos A + cos B = 2 cos 2 A + B cos 2 A − B ,
LHS = sin 5 x + sin 3 x cos 5 x + cos 3 x = 2 sin 5 x + 3 x 2 cos 5 x − 3 x 2 2 cos 5 x + 3 x 2 cos 5 x − 3 x 2 = 2 sin 4 x cos x 2 cos 4 x cos x = sin 4 x cos 4 x = tan 4 x = RHS
\begin{array}{l}
\displaystyle \text{LHS}=\frac{\sin5x+\sin3x}{\cos5x+\cos3x} \\[1em]
\hspace{1.6em} \displaystyle =\frac{2\sin\dfrac{5x+3x}{2}\cos\dfrac{5x-3x}{2}}{2\cos\dfrac{5x+3x}{2}\cos\dfrac{5x-3x}{2}} \\[1.8em]
\hspace{1.6em} \displaystyle =\frac{2\sin4x\cos x}{2\cos4x\cos x} \\[1em]
\hspace{1.6em} \displaystyle =\frac{\sin4x}{\cos4x} \\[1em]
\hspace{1.6em} \displaystyle =\tan4x=\text{RHS}
\end{array}
LHS = cos 5 x + cos 3 x sin 5 x + sin 3 x = 2 cos 2 5 x + 3 x cos 2 5 x − 3 x 2 sin 2 5 x + 3 x cos 2 5 x − 3 x = 2 cos 4 x cos x 2 sin 4 x cos x = cos 4 x sin 4 x = tan 4 x = RHS Hence proved.
Q18. Prove the following: sin x − sin y cos x + cos y = tan x − y 2 \frac{\sin x-\sin y}{\cos x+\cos y}=\tan\frac{x-y}{2} c o s x + c o s y s i n x − s i n y = tan 2 x − y
Sol. Using sin A − sin B = 2 cos A + B 2 sin A − B 2 \sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2} sin A − sin B = 2 cos 2 A + B sin 2 A − B
and cos A + cos B = 2 cos A + B 2 cos A − B 2 \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2} cos A + cos B = 2 cos 2 A + B cos 2 A − B ,
LHS = sin x − sin y cos x + cos y = 2 cos x + y 2 sin x − y 2 2 cos x + y 2 cos x − y 2
\begin{array}{l}
\displaystyle \text{LHS}=\frac{\sin x-\sin y}{\cos x+\cos y} \\[1em]
\hspace{1.8em} \displaystyle =\frac{2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}}{2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}}
\end{array}
LHS = cos x + cos y sin x − sin y = 2 cos 2 x + y cos 2 x − y 2 cos 2 x + y sin 2 x − y = sin x − y 2 cos x − y 2 = tan x − y 2 = RHS
\begin{array}{l}
\hspace{1.8em} \displaystyle =\frac{\sin\dfrac{x-y}{2}}{\cos\dfrac{x-y}{2}} \\[1.8em]
\hspace{1.8em} \displaystyle =\tan\frac{x-y}{2}=\text{RHS}
\end{array}
= cos 2 x − y sin 2 x − y = tan 2 x − y = RHS Hence proved.
Q19. Prove the following: sin x + sin 3 x cos x + cos 3 x = tan 2 x \frac{\sin x+\sin3x}{\cos x+\cos3x}=\tan2x c o s x + c o s 3 x s i n x + s i n 3 x = tan 2 x
Sol. Using sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B
and cos A + cos B = 2 cos A + B 2 cos A − B 2 \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2} cos A + cos B = 2 cos 2 A + B cos 2 A − B ,
LHS = sin x + sin 3 x cos x + cos 3 x = 2 sin x + 3 x 2 cos x − 3 x 2 2 cos x + 3 x 2 cos x − 3 x 2 = 2 sin 2 x cos ( − x ) 2 cos 2 x cos ( − x ) = sin 2 x cos 2 x = tan 2 x = RHS
\begin{array}{l}
\displaystyle \text{LHS}=\frac{\sin x+\sin3x}{\cos x+\cos3x} \\[1.2em]
\hspace{1.6em} \displaystyle =\frac{2\sin\dfrac{x+3x}{2}\cos\dfrac{x-3x}{2}}{2\cos\dfrac{x+3x}{2}\cos\dfrac{x-3x}{2}} \\[1.2em]
\hspace{1.6em} \displaystyle =\frac{2\sin2x\cos(-x)}{2\cos2x\cos(-x)} \\[1.2em]
\hspace{1.6em} \displaystyle =\frac{\sin2x}{\cos2x} \\[0.8em]
\hspace{1.6em} \displaystyle =\tan2x=\text{RHS}
\end{array}
LHS = cos x + cos 3 x sin x + sin 3 x = 2 cos 2 x + 3 x cos 2 x − 3 x 2 sin 2 x + 3 x cos 2 x − 3 x = 2 cos 2 x cos ( − x ) 2 sin 2 x cos ( − x ) = cos 2 x sin 2 x = tan 2 x = RHS Hence proved.
Q20. Prove the following: sin x − sin 3 x sin 2 x − cos 2 x = 2 sin x \frac{\sin x-\sin3x}{\sin^2x-\cos^2x}=2\sin x s i n 2 x − c o s 2 x s i n x − s i n 3 x = 2 sin x
Sol.
LHS = sin x − sin 3 x sin 2 x − cos 2 x
\text{LHS}=\frac{\sin x-\sin3x}{\sin^2x-\cos^2x}
LHS = sin 2 x − cos 2 x sin x − sin 3 x Taking − 1 -1 − 1 common from the numerator and from the denominator,
= − ( sin 3 x − sin x ) − ( cos 2 x − sin 2 x ) = sin 3 x − sin x cos 2 x − sin 2 x
\begin{array}{l}
\hspace{1em} \displaystyle =\frac{-(\sin3x-\sin x)}{-(\cos^2x-\sin^2x)} \\[1em]
\hspace{1em} \displaystyle =\frac{\sin3x-\sin x}{\cos^2x-\sin^2x}
\end{array}
= − ( cos 2 x − sin 2 x ) − ( sin 3 x − sin x ) = cos 2 x − sin 2 x sin 3 x − sin x Using sin A − sin B = 2 cos A + B 2 sin A − B 2 \sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2} sin A − sin B = 2 cos 2 A + B sin 2 A − B and cos 2 x − sin 2 x = cos 2 x \cos^2x-\sin^2x=\cos2x cos 2 x − sin 2 x = cos 2 x ,
= 2 cos 3 x + x 2 sin 3 x − x 2 cos 2 x = 2 cos 2 x sin x cos 2 x = 2 sin x = RHS
\begin{array}{l}
\hspace{1em} \displaystyle =\frac{2\cos\dfrac{3x+x}{2}\sin\dfrac{3x-x}{2}}{\cos2x} \\[1em]
\hspace{1em} \displaystyle =\frac{2\cos2x\sin x}{\cos2x} \\[1em]
\hspace{1em} \displaystyle =2\sin x=\text{RHS}
\end{array}
= cos 2 x 2 cos 2 3 x + x sin 2 3 x − x = cos 2 x 2 cos 2 x sin x = 2 sin x = RHS Hence proved.
Q21. Prove the following: cos 4 x + cos 3 x + cos 2 x sin 4 x + sin 3 x + sin 2 x = cot 3 x \frac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x}=\cot3x s i n 4 x + s i n 3 x + s i n 2 x c o s 4 x + c o s 3 x + c o s 2 x = cot 3 x
Sol.
LHS = ( cos 4 x + cos 2 x ) + cos 3 x ( sin 4 x + sin 2 x ) + sin 3 x
\text{LHS}=\frac{(\cos4x+\cos2x)+\cos3x}{(\sin4x+\sin2x)+\sin3x}
LHS = ( sin 4 x + sin 2 x ) + sin 3 x ( cos 4 x + cos 2 x ) + cos 3 x Using cos A + cos B = 2 cos A + B 2 cos A − B 2 \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2} cos A + cos B = 2 cos 2 A + B cos 2 A − B
and sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B ,
= 2 cos 4 x + 2 x 2 cos 4 x − 2 x 2 + cos 3 x 2 sin 4 x + 2 x 2 cos 4 x − 2 x 2 + sin 3 x = 2 cos 3 x cos x + cos 3 x 2 sin 3 x cos x + sin 3 x
\begin{array}{l}
\hspace{1.2em} \displaystyle =\frac{2\cos\dfrac{4x+2x}{2}\cos\dfrac{4x-2x}{2}+\cos3x}{2\sin\dfrac{4x+2x}{2}\cos\dfrac{4x-2x}{2}+\sin3x} \\[1.2em]
\hspace{1.2em} \displaystyle =\frac{2\cos3x\cos x+\cos3x}{2\sin3x\cos x+\sin3x}
\end{array}
= 2 sin 2 4 x + 2 x cos 2 4 x − 2 x + sin 3 x 2 cos 2 4 x + 2 x cos 2 4 x − 2 x + cos 3 x = 2 sin 3 x cos x + sin 3 x 2 cos 3 x cos x + cos 3 x = cos 3 x ( 2 cos x + 1 ) sin 3 x ( 2 cos x + 1 ) = cos 3 x sin 3 x = cot 3 x = RHS
\begin{array}{l}
\hspace{1.2em} \displaystyle =\frac{\cos3x(2\cos x+1)}{\sin3x(2\cos x+1)} \\[1em]
\hspace{1.2em} \displaystyle =\frac{\cos3x}{\sin3x} \\[0.8em]
\hspace{1.2em} \displaystyle =\cot3x=\text{RHS}
\end{array}
= sin 3 x ( 2 cos x + 1 ) cos 3 x ( 2 cos x + 1 ) = sin 3 x cos 3 x = cot 3 x = RHS Hence proved.
Q22. Prove the following: cot x cot 2 x − cot 2 x cot 3 x − cot 3 x cot x = 1 \cot x\cot2x-\cot2x\cot3x-\cot3x\cot x=1 cot x cot 2 x − cot 2 x cot 3 x − cot 3 x cot x = 1
Sol. We know that 3 x = 2 x + x 3x=2x+x 3 x = 2 x + x .
cot 3 x = cot ( 2 x + x )
\cot3x=\cot(2x+x)
cot 3 x = cot ( 2 x + x ) Using cot ( A + B ) = cot A cot B − 1 cot B + cot A \cot(A+B)=\frac{\cot A\cot B-1}{\cot B+\cot A} cot ( A + B ) = c o t B + c o t A c o t A c o t B − 1 ,
cot 3 x = cot 2 x cot x − 1 cot x + cot 2 x
\cot3x=\frac{\cot2x\cot x-1}{\cot x+\cot2x}
cot 3 x = cot x + cot 2 x cot 2 x cot x − 1 Cross-multiplying,
cot 3 x ( cot x + cot 2 x ) = cot 2 x cot x − 1 cot 3 x cot x + cot 3 x cot 2 x = cot 2 x cot x − 1
\begin{array}{l}
\displaystyle \cot3x(\cot x+\cot2x)=\cot2x\cot x-1 \\[0.4em]
\displaystyle \cot3x\cot x+\cot3x\cot2x=\cot2x\cot x-1
\end{array}
cot 3 x ( cot x + cot 2 x ) = cot 2 x cot x − 1 cot 3 x cot x + cot 3 x cot 2 x = cot 2 x cot x − 1 cot x cot 2 x − cot 2 x cot 3 x − cot 3 x cot x = 1
\cot x\cot2x-\cot2x\cot3x-\cot3x\cot x=1
cot x cot 2 x − cot 2 x cot 3 x − cot 3 x cot x = 1 Hence proved.
Q23. Prove the following: tan 4 x = 4 tan x ( 1 − tan 2 x ) 1 − 6 tan 2 x + tan 4 x \tan4x=\frac{4\tan x(1-\tan^2x)}{1-6\tan^2x+\tan^4x} tan 4 x = 1 − 6 t a n 2 x + t a n 4 x 4 t a n x ( 1 − t a n 2 x )
Sol.
LHS = tan 4 x = tan 2 ( 2 x )
\begin{array}{l}
\displaystyle \text{LHS}=\tan4x \\[0.4em]
\hspace{0.2em} \displaystyle =\tan2(2x)
\end{array}
LHS = tan 4 x = tan 2 ( 2 x ) Using tan 2 A = 2 tan A 1 − tan 2 A \tan2A=\frac{2\tan A}{1-\tan^2A} tan 2 A = 1 − t a n 2 A 2 t a n A ,
= 2 tan 2 x 1 − tan 2 ( 2 x )
\begin{array}{l}
\hspace{0.2em} \displaystyle =\frac{2\tan2x}{1-\tan^2(2x)}
\end{array}
= 1 − tan 2 ( 2 x ) 2 tan 2 x Using tan 2 x = 2 tan x 1 − tan 2 x \tan2x=\frac{2\tan x}{1-\tan^2x} tan 2 x = 1 − t a n 2 x 2 t a n x ,
= 2 × 2 tan x 1 − tan 2 x 1 − ( 2 tan x 1 − tan 2 x ) 2
\begin{array}{l}
\hspace{0.2em} \displaystyle =\frac{2\times\dfrac{2\tan x}{1-\tan^2x}}{1-\left(\dfrac{2\tan x}{1-\tan^2x}\right)^2}
\end{array}
= 1 − ( 1 − tan 2 x 2 tan x ) 2 2 × 1 − tan 2 x 2 tan x = 4 tan x 1 − tan 2 x ( 1 − tan 2 x ) 2 − 4 tan 2 x ( 1 − tan 2 x ) 2 = 4 tan x ( 1 − tan 2 x ) ( 1 − tan 2 x ) 2 − 4 tan 2 x = 4 tan x ( 1 − tan 2 x ) 1 − 2 tan 2 x + tan 4 x − 4 tan 2 x = 4 tan x ( 1 − tan 2 x ) 1 − 6 tan 2 x + tan 4 x = RHS
\begin{array}{l}
\hspace{0.2em} \displaystyle =\frac{\dfrac{4\tan x}{1-\tan^2x}}{\dfrac{(1-\tan^2x)^2-4\tan^2x}{(1-\tan^2x)^2}} \\[2em]
\hspace{0.2em} \displaystyle =\frac{4\tan x(1-\tan^2x)}{(1-\tan^2x)^2-4\tan^2x} \\[1em]
\hspace{0.2em} \displaystyle =\frac{4\tan x(1-\tan^2x)}{1-2\tan^2x+\tan^4x-4\tan^2x} \\[1em]
\hspace{0.2em} \displaystyle =\frac{4\tan x(1-\tan^2x)}{1-6\tan^2x+\tan^4x}=\text{RHS}
\end{array}
= ( 1 − tan 2 x ) 2 ( 1 − tan 2 x ) 2 − 4 tan 2 x 1 − tan 2 x 4 tan x = ( 1 − tan 2 x ) 2 − 4 tan 2 x 4 tan x ( 1 − tan 2 x ) = 1 − 2 tan 2 x + tan 4 x − 4 tan 2 x 4 tan x ( 1 − tan 2 x ) = 1 − 6 tan 2 x + tan 4 x 4 tan x ( 1 − tan 2 x ) = RHS Hence proved.
Q24. Prove the following: cos 4 x = 1 − 8 sin 2 x cos 2 x \cos4x=1-8\sin^2x\cos^2x cos 4 x = 1 − 8 sin 2 x cos 2 x
Sol.
LHS = cos 4 x = cos 2 ( 2 x )
\begin{array}{l}
\displaystyle \text{LHS}=\cos4x \\[0.4em]
\hspace{0.8em} \displaystyle =\cos2(2x)
\end{array}
LHS = cos 4 x = cos 2 ( 2 x ) Using cos 2 A = 1 − 2 sin 2 A \cos2A=1-2\sin^2A cos 2 A = 1 − 2 sin 2 A ,
= 1 − 2 sin 2 2 x
\begin{array}{l}
\hspace{0.8em} \displaystyle =1-2\sin^22x
\end{array}
= 1 − 2 sin 2 2 x Using sin 2 A = 2 sin A cos A \sin2A=2\sin A\cos A sin 2 A = 2 sin A cos A ,
= 1 − 2 ( 2 sin x cos x ) 2
\begin{array}{l}
\hspace{0.8em} \displaystyle =1-2(2\sin x\cos x)^2
\end{array}
= 1 − 2 ( 2 sin x cos x ) 2 = 1 − 2 × 4 sin 2 x cos 2 x = 1 − 8 sin 2 x cos 2 x = RHS
\begin{array}{l}
\hspace{0.8em} \displaystyle =1-2\times4\sin^2x\cos^2x \\[0.4em]
\hspace{0.8em} \displaystyle =1-8\sin^2x\cos^2x=\text{RHS}
\end{array}
= 1 − 2 × 4 sin 2 x cos 2 x = 1 − 8 sin 2 x cos 2 x = RHS Hence proved.
Q25. Prove the following: cos 6 x = 32 cos 6 x − 48 cos 4 x + 18 cos 2 x − 1 \cos6x=32\cos^6x-48\cos^4x+18\cos^2x-1 cos 6 x = 32 cos 6 x − 48 cos 4 x + 18 cos 2 x − 1
Sol.
LHS = cos 6 x = cos 3 ( 2 x )
\begin{array}{l}
\displaystyle \text{LHS}=\cos6x \\[0.4em]
\hspace{0.8em} \displaystyle =\cos3(2x)
\end{array}
LHS = cos 6 x = cos 3 ( 2 x ) Using cos 3 A = 4 cos 3 A − 3 cos A \cos3A=4\cos^3A-3\cos A cos 3 A = 4 cos 3 A − 3 cos A ,
= 4 cos 3 2 x − 3 cos 2 x
\begin{array}{l}
\hspace{0.8em} \displaystyle =4\cos^32x-3\cos2x
\end{array}
= 4 cos 3 2 x − 3 cos 2 x Using cos 2 x = 2 cos 2 x − 1 \cos2x=2\cos^2x-1 cos 2 x = 2 cos 2 x − 1 ,
= 4 ( 2 cos 2 x − 1 ) 3 − 3 ( 2 cos 2 x − 1 )
\begin{array}{l}
\hspace{0.8em} \displaystyle =4(2\cos^2x-1)^3-3(2\cos^2x-1)
\end{array}
= 4 ( 2 cos 2 x − 1 ) 3 − 3 ( 2 cos 2 x − 1 ) Using ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 (a-b)^3=a^3-3a^2b+3ab^2-b^3 ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 ,
= 4 [ ( 2 cos 2 x ) 3 − ( 1 ) 3 − 3 ( 2 cos 2 x ) 2 + 3 ( 2 cos 2 x ) ] − 6 cos 2 x + 3
\begin{array}{l}
\hspace{0.8em} \displaystyle =4\left[(2\cos^2x)^3-(1)^3-3(2\cos^2x)^2+3(2\cos^2x)\right]-6\cos^2x+3
\end{array}
= 4 [ ( 2 cos 2 x ) 3 − ( 1 ) 3 − 3 ( 2 cos 2 x ) 2 + 3 ( 2 cos 2 x ) ] − 6 cos 2 x + 3 = 4 [ 8 cos 6 x − 1 − 12 cos 4 x + 6 cos 2 x ] − 6 cos 2 x + 3 = 32 cos 6 x − 4 − 48 cos 4 x + 24 cos 2 x − 6 cos 2 x + 3 = 32 cos 6 x − 48 cos 4 x + 18 cos 2 x − 1 = RHS
\begin{array}{l}
\hspace{0.8em} \displaystyle =4\left[8\cos^6x-1-12\cos^4x+6\cos^2x\right]-6\cos^2x+3 \\[0.4em]
\hspace{0.8em} \displaystyle =32\cos^6x-4-48\cos^4x+24\cos^2x-6\cos^2x+3 \\[0.4em]
\hspace{0.8em} \displaystyle =32\cos^6x-48\cos^4x+18\cos^2x-1=\text{RHS}
\end{array}
= 4 [ 8 cos 6 x − 1 − 12 cos 4 x + 6 cos 2 x ] − 6 cos 2 x + 3 = 32 cos 6 x − 4 − 48 cos 4 x + 24 cos 2 x − 6 cos 2 x + 3 = 32 cos 6 x − 48 cos 4 x + 18 cos 2 x − 1 = RHS Hence proved.