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{2 x}{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 $