Question

If $\varphi$ is an acute angle such that $\tan \varphi =\frac{2}{3}$, then evaluate
$\left(\frac{1+\tan \varphi }{\sin \varphi +\cos \varphi }\right)\left(\frac{1-\cot \varphi }{\sec \varphi +\text{cosec}\varphi }\right)$

• A. $\frac{-1}{5}$
• B. $\frac{-4}{\sqrt{13}}$
• C. $\frac{1}{5}$
• D. $\frac{4}{\sqrt{13}}$

Answer 1

The correct option is C $\frac{1}{5}$

Given : $\tan \varphi =\frac{2}{3}$

Consider, $\left(\frac{1+\tan \varphi }{\sin \varphi +\cos \varphi }\right)\left(\frac{1-\cot \varphi }{\sec \varphi +\text{cosec}\varphi }\right)$

$=\left(\frac{1+\tan \varphi }{\sin \varphi +\cos \varphi }\right)\left(\frac{1-\frac{1}{\tan \varphi }}{\frac{1}{\cos \varphi }+\frac{1}{\sin \varphi }}\right)$

$=-\left(\frac{(1+\tan \varphi )(1-\tan \varphi )}{(\frac{\sin \varphi }{\cos \varphi }+\frac{\cos \varphi }{\sin \varphi }+1+1)\tan \varphi }\right)$

$=-\left(\frac{(1+\tan \varphi )(1-\tan \varphi )}{(\tan \varphi +\frac{1}{\tan \varphi }+2)\tan \varphi }\right)$

$=-\left(\frac{(1+\tan \varphi )(1-\tan \varphi )}{\frac{(\tan \varphi +1{)}^2}{\tan \varphi }\tan \varphi }\right)$

$=-\frac{(1-\tan \varphi )}{(\tan \varphi +1)}$

$=-\frac{\frac{2}{3}-1}{\frac{2}{3}+1}$

$=\frac{1}{5}$.

Hence, option C is correct.