Question

$tan(\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}x)+tan(\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}x),x\ne 0$, is equal to

• A. $x$
• B. $2x$
• C. $\frac{2}{x}$
• D. $noneofthese$

Answer 1

The correct option is C $\frac{2}{x}$

Let $\frac{1}{2}co{s}^{-1}\left(x\right)=A$
Therefore simplifying the above expression, we get
$\frac{1-tanA}{1+tanA}+\frac{1+tanA}{1-tanA}$

$=\frac{2(1+ta{n}^{2}A)}{1-ta{n}^{2}A}$

$=\frac{2}{cos2A}$

$=\frac{2}{cos\left(co{s}^{-1}\right(x\left)\right)}$

$=\frac{2}{x}$