Prove that: (3 2 +x ) (2 +x) [ (3 2 -x )+ (2 +x) ]=1

Question

Prove that: $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$

✓

Answer

To prove: $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
Taking L.H.S., we have
LHS = $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
$\cos \left(n \frac{\pi}{2}+\theta\right) = \sin \theta$, if $n$ is odd
$\cos \left(n \frac{\pi}{2}+\theta\right)=\cos \theta$ if $n$ is even we know that,
$\cot \left(n \frac{\pi}{2}-\theta\right)=\tan \theta$ if $n$ is odd
$\cot \left(n \frac{\pi}{2}+\theta\right) = \cot \theta$, if $n$ is even
Therefore, L.H.S becomes,
LHS $= \sin x \cos x [\tan x+\cot x] = \sin x \cos x \left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]$
L.H.S $= \sin x. \sin x + \cos x. \cos x$
L.H.S $= \sin^2 x + \cos^2 x$
Since, $\sin^2 x + \cos^2 x = 1$
So, LHS $= 1$
$\therefore$ LHS = RHS

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "2 Marks Questions" from Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Trigonometric Functions — Maths STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions