Question

The value of $\sec \left(2{\cot }^{-1}2+{\cos }^{-1}\frac{3}{5}\right)$ is equal to

• A. $\frac{25}{24}$
• B. $-\frac{24}{7}$
• C. $\frac{25}{7}$
• D. $-\frac{25}{7}$

Answer 1

Answer: (D)

$\alpha =2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{4}{3}$
$={\tan }^{-1}\frac{1}{1-\frac{1}{4}}+{\tan }^{-1}\frac{4}{3}=2{\tan }^{-1}\frac{4}{3}\text{ (using }2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}\text{ ) }$
$\text{ now }\sec (2\alpha )\text{ where }\tan \alpha =4/3$
$\frac{1+{\tan }^2\theta }{1-{\tan }^2\theta }=\frac{1+(16/9)}{1-(16/9)}=\frac{25}{-7}$
$\therefore -\sec (\theta )\text{ where }\tan \theta =\frac{24}{7}.\text{ Hence }\sec \theta =-\frac{25}{7}\Rightarrow D$