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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 25 of 61
Marks: +1, -0
cos⁡(π+x)cos⁡(−x)sin⁡(π−x)cos⁡(π2+x)\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\frac{\pi}{2}+x\right)} = cot⁡2\cot^2 x
Solution:  
We have, L.H.S. = cos⁡(π+x)cos⁡(−x)sin⁡(π−x)cos⁡(π2+x)\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\frac{\pi}{2}+x\right)}
= −cos⁡xcos⁡xsin⁡x(−sin⁡x)\frac{-\cos x \cos x}{\sin x (-\sin x)} = cot⁡2\cot^2 x = R.H.S.
[Since cos (π+ θ) = - cos θ , cos (π2+θ)\left(\frac{\pi}{2} + \theta\right) = - sin θ
cos (- θ) = cos θ , sin (π - θ) = sinθ]
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