Inverse Trigonometric Functions

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Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

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Kindly go through the solution

= tan ^-1 $\frac{sec-\theta }{tan\theta }$

tan ^-1 $\frac{\frac{1}{cos\theta }-1}{\frac{sin\theta }{cos\theta }}$ = tan ^-1 $\frac{\frac{1-cos\theta }{sin\theta }}{\frac{sin\theta }{cos\theta }}$ $=ta{n}^{+}\frac{1-cos\theta }{sin\theta }=ta{n}^{-1}\frac{2si{n}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=la{n}^{-1}\frac{sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ { Ø cos 2Ø = 1 - sin²ØÞ 2 sin2 Ø = 1 - cos 2Ø => 2 sin2 $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ = 1 cos2 $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ = 1- cosØ }

$=ta{n}^{-1}\left(tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ Similar

$=\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ sin 2Ø = 2sinØ cosØ => sin $\frac{\theta }{2}$ = 2 sin $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ cos $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ => sin = 2 sin $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ cos $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$=\frac{ta{n}^{-1}x}{2}$

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