Question

$\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]+\cot [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]$ is equal to

• A. $\frac{5}{7}$
• B. $\frac{10}{7}$
• C. $\frac{14}{5}$
• D. $\frac{7}{5}$

Answer 1

The correct option is C $\frac{14}{5}$

$\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]+\cot [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]$

$=\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]$

We are using the formula:

$\tan [\frac{\pi }{4}+\frac{x}{2}]+\tan [\frac{\pi }{4}-\frac{x}{2}]=\frac{2}{\cos x},x={\cos }^{-1}\frac{a}{b}$

$\tan [\frac{\pi }{4}+\frac{x}{2}]+\tan [\frac{\pi }{4}-\frac{x}{2}]=\frac{2b}{a}$

$\therefore \tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]+\cot [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]=\frac{2\times 7}{5}$

$\Rightarrow \therefore \tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]+\cot [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\left(\frac{5}{7}\right)]=\frac{14}{5}$