Question

$2{\tan }^{-1}(\cos x)={\tan }^{-1}\left(\cos e{c}^{2}x\right),thenx=$

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\mathrm{\pi }$
• C. $\frac{\mathrm{\pi }}{6}$
• D. $\frac{\mathrm{\pi }}{3}$

Answer 1

The correct option is D

$\frac{\mathrm{\pi }}{3}$

Find the value of given equation:

$2{\tan }^{-1}(\cos x)={\tan }^{-1}\left(\cos e{c}^{2}x\right)$

$\Rightarrow$ $\frac{2\cos x}{1-{\cos }^{2}x}=\frac{1}{{\sin }^{2}x}$ $\left[\mathbf{\because }\mathbf{2}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{=}\mathbf{\left(}\frac{\mathbf{2}\mathbf{x}}{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{\right)}\right]$

$\Rightarrow$ $\frac{2\cos x}{{\sin }^{2}x}=\frac{1}{{\sin }^{2}x}$ $\left[\mathbf{\because }\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\right]$

$\Rightarrow$ $2\cos x=1$

$\Rightarrow$ $\cos x=\frac{1}{2}$

$\Rightarrow$ $x=\frac{\mathbf{\pi }}{\mathbf{3}}$ $\left[\mathbf{\because }\mathbf{-}\mathbf{1}\mathbf{\le }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{}\mathbf{\le }\mathbf{1}\right]$

Hence, Option ‘D’ is Correct.