Question

prove that 2tan-1(cosx)=tan-1(2cosecx)

Answer 1

$using \: \\ tan {}^{ - 1} x + {tan}^{ - 1} y \\ = {tan}^{ - 1} ( \frac{x + y}{1 - xy} )$
here x=y =cosx
so., LHS becomes
$\: \: \: \: \: 2 {tan}^{ - 1} (cosx) \\ = {tan}^{ - 1}(cosx) + {tan}^{ - 1} (cosx) \\ = {tan}^{ - 1}( \frac{2cosx}{1 - {cos}^{2} x} ) \\ = {tan}^{ - 1} ( \frac{2cosx}{sin {}^{2}x } ) \\ = {tan}^{ - 1} (2cosecx)$

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Answer 2

To prove that: 2$\tan^{-1}(\cos x)$ = $\tan^{-1}(2\csc x)$.

Solution:

We know that,

$2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}$

Put x = $\cos x$, we get

2$\tan^{-1}(\cos x)$ = $\tan^{-1}\dfrac{2\cos x}{1-\cos^2 x}$

⇒ 2$\tan^{-1}(\cos x)$ = $\tan^{-1}\dfrac{2\cos x}{\sin^2 x}$              ........... (1)

[∵ $\sin^2 A+\cos^2 A$ = 1]

Given by question,

2$\tan^{-1}(\cos x)$ = $\tan^{-1}(2\csc x)$                   ........... (2)

Comparing equations (1) and (2), we get

$\dfrac{2\cos x}{\sin^2 x}$ = 2$\csc x$

⇒  $\dfrac{\cos x}{\sin^2 x}$ = $\dfrac{1}{\sin x}$

⇒ $\tan x$ =

⇒ $\tan x$ = $\tan 45$°

⇒ x = 45°

∴ 2$\tan^{-1}(\cos x)$ = $\tan^{-1}(2\csc x)$, proved.

Thus, 2$\tan^{-1}(\cos x)$ = $\tan^{-1}(2\csc x)$, proved.