Question

If $y=\left(t a n\right)^{- 1}\left(s e c x - t a n x\right),$ then the value of $\frac{d y}{d x}$ is

Answer 1

$y=\left(t a n\right)^{- 1}\left(s e c x - t a n x\right)$
$y=\left(t a n\right)^{- 1}\left(\frac{1 - s i n x}{c o s x}\right)$
$\Rightarrow y=(t a n)^{- 1}(\frac{(c o s \frac{x}{2} - s i n \frac{x}{2})^{2}}{(c o s)^{2} \frac{x}{2} - (s i n)^{2} \frac{x}{2})})$
$\Rightarrow y=\left(t a n\right)^{- 1}\left(\frac{c o s \frac{x}{2} - s i n \frac{x}{2}}{c o s \frac{x}{2} + s i n \frac{x}{2}}\right)$
$\Rightarrow y=\left(t a n\right)^{- 1}\left(\frac{1 - t a n \frac{x}{2}}{1 + t a n \frac{x}{2}}\right)$
$\Rightarrow y=(t a n)^{- 1}(t a n (\frac{\pi }{4} - \frac{x}{2}))$
$\Rightarrow y=\frac{\pi }{4}-\frac{x}{2}$
Differentiating with respect to $x,$ we get
$\Rightarrow \frac{d y}{d x}=\frac{- 1}{2}=-0.5$