Question

If $2 \,tan^{-1} (cos\,x) = tan^{-1} (2 \,cosec\,x)$ then $sin\,x + cos\,x$ is equal to

• A. $2 \sqrt2$
• B. $\sqrt2$
• C. $\frac {1}{\sqrt2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\sqrt2$

Answer 2

Given, $2\,tan^{-1} (cos\,x) = tan^{-1} (2 \,cosec\,x)$
$\Rightarrow tan ^{-1} \frac{2 \,cos \,x}{1-cos ^{2} \,x}=\tan ^{-1}\left(\frac{2}{sin \,x}\right)$
$\Rightarrow \frac{2\,cos\, x}{1- cos ^{2} x}=\frac{2}{\sin \,x}$
$\Rightarrow \frac{cos \,x}{sin ^{2} \,x}=\frac{1}{sin \,x}$
$\Rightarrow \frac{cos\, x}{sin\, x}=1$
$[\because sin x=0]$
$\Rightarrow tan \, x=1$
$\Rightarrow x = \frac{\pi}{4}$
Now, $sin\,x + cos\,x = sin \frac{\pi}{4} + cos \frac{\pi}{4}$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$