Maths Inverse Trigonometric Functions Inverse Trigonometric Functions Class 12th Ncert Solutions Maths class 12th

Kindly Consider the following

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Vishal Baghel | Contributor-Level 10

8 months ago

Kindly go through the solution

= tan ^-1 $\frac{s e c - θ}{t a n θ}$

tan ^-1 $\frac{\frac{1}{c o s θ} - 1}{\frac{s i n θ}{c o s θ}}$ = tan ^-1 $\frac{\frac{1 - c o s θ}{s i n θ}}{\frac{s i n θ}{c o s θ}}$ $= t a n^{+} \frac{1 - c o s θ}{s i n θ} = t a n^{- 1} \frac{2 s i n^{2} \frac{θ}{2}}{2 s i n \frac{θ}{2} c o s \frac{θ}{2}}$

$= l a n^{- 1} \frac{s i n \frac{θ}{2}}{c o s \frac{θ}{2}}$ { Ø cos 2Ø = 1 - sin²ØÞ 2 sin2 Ø = 1 - cos 2Ø => 2 sin2 $\frac{θ}{2}$ = 1 cos2 $\frac{θ}{2}$ = 1- cosØ }

$= t a n^{- 1} (t a n \frac{θ}{2})$ Similar

$= \frac{θ}{2}$ sin 2Ø = 2sinØ cosØ => sin $\frac{θ}{2}$ = 2 sin $\frac{θ}{2}$ cos $\frac{θ}{2}$ => sin = 2 sin $\frac{θ}{2}$ cos $\frac{θ}{2}$

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

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Maths Inverse Trigonometric Functions 2021

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