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

Answers (1)

We have $\cos \left ( \tan^{-1}x \right )=\sin\left ( \cot^{-1}\frac{3}{4} \right )$

$\Rightarrow \cos \left ( \cos^{-1}\frac{1}{\sqrt{x^{2}+2}} \right )=\sin\left ( \sin^{-1}\frac{4}{5} \right )........(i)$

Let $\tan^{-1}x=\theta _{1}\Rightarrow \tan\theta_{1}=\frac{x}{1}$

$\Rightarrow \cos \theta_{1}=\frac{1}{\sqrt{x^{2}+1}}.....(a)$

$\Rightarrow \theta_{1}=\cos^{-1}\frac{1}{\sqrt{x^{2}+1}}.....(c)$

And $\cot^{-1}=\theta_{2}\Rightarrow \cot^{-1}=\frac{3}{4}$

$\Rightarrow \sin \theta_{2}=\frac{4}{5}.......(b)$

$\Rightarrow \theta_{2}= \sin^{-1}\frac{4}{5}.......(d)$

From (c),(d);(i) becomes

$\Rightarrow \cos \theta_{1}= \sin\theta_{2}$

$\Rightarrow \frac{1}{\sqrt{x^{2}+1}}=\frac{4}{5}$ [From (a),(b)]

On squarinting both Sides, we get

$\Rightarrow 16\left (x^{2}+1 \right )=25$

$\Rightarrow 16x^{2}=9$

$\Rightarrow x^{2}=\left (\frac{3}{4} \right )^{2}$

$\Rightarrow x=\pm \frac{3}{4}$

$\therefore x=\frac{3}{4},-\frac{3}{4}.$

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infoexpert24

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