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Questions  >>  CBSE XII  >>  Math  >>  Model Papers
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Q)

Solve the following equation :$\cos\big(tan^{-1}x\big)=sin\bigg(cot^{-1}\frac{3}{4}\bigg)$

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

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A)
Toolbox:
  • $cos (x) = sin(\frac{\pi}{2} - x)$
  • $cot^{-1}x = \frac{\pi}{2}-tan^{-1}x$
Given $cos\big(tan^{-1}x\big)=sin\big(cot^{-1}\frac{3}{4}\big)$
We know that $cos (x) = sin(\frac{\pi}{2} - x)$
$\Rightarrow sin\big(\frac{\pi}{2}-\tan^{-1}x\big)=sin\big(cot^{-1}\frac{3}{4}\big)$
$\Rightarrow \frac{\pi}{2}-tan^{-1}x=cot^{-1}\frac{3}{4}$
We know that $\frac{\pi}{2}-tan^{-1}x=cot^{-1}x$
$\Rightarrow cot^{-1}x=cot^{-1}\frac{3}{4}$
$\Rightarrow x=\frac{3}{4}$
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