Question

If $2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left(2\text{cosec}x\right)$, then $\sin x+\cos x=$

• A. $2\sqrt{2}$
• B. $\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\sqrt{2}$

${\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

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

$\Rightarrow$ ${\tan }^{-1}\left(\frac{2\cos x}{1-{\cos }^{2}x}\right)={\tan }^{-1}\left(\frac{2}{\sin x}\right)$ $\because$(${\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$)

$\Rightarrow$ $\frac{2\cos x}{{\sin }^{2}x}=\frac{2}{\sin x}$

$\Rightarrow$$\cos x=\sin x$

$\Rightarrow$$x={45}^{\circ }$

$\therefore$ $\cos x+\sin x=2\sin x=2\sin \left({45}^{\circ }\right)=2\times \frac{1}{\sqrt{2}}=\sqrt{2}$