Tan -1-cos x -1-sin x - A- -4-x-2 - -4-x-2 C- x-2 D- -4-x

The correct option is A tan^ 1[cos x/1+sin x ]=tan^ 1[sin(π/2 x)/1+cos(π/2 x) ] =tan^ 1[2 sin(π/4)cos(π/4 x/2)/2 cos^2(π/4 x/2) ] =tan^ 1tan(π/4 x/2 )=π/4 x/2. ...

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Byju's Answer

Standard XII

Mathematics

Composition of Inverse Trigonometric Functions and Trigonometric Functions

tan -1[cos x ...

Question

$tan^{-1}\left[\frac{cos~x}{1+sin~x} \right ]=$

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Solution

The correct option is A
$t a n^{- 1} [\frac{c o s x}{1 + s i n x}] = t a n^{- 1} [\frac{s i n (π / 2 - x)}{1 + c o s (π / 2 - x)}]$
$= t a n^{- 1} [\frac{2 s i n (π / 4) c o s (π / 4 - x / 2)}{2 c o s^{2} (π / 4 - x / 2)}]$
$= t a n^{- 1} t a n (\frac{π}{4} - \frac{x}{2}) = \frac{π}{4} - \frac{x}{2} .$

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\(tan^{-1}\left[\frac{cos~x}{1+sin~x} \right ]=\)

The correct option is A tan−1[cos x1+sin x]=tan−1[sin(π/2−x)1+cos(π/2−x)] =tan−1[2 sin(π/4)cos(π/4−x/2)2 cos2(π/4−x/2)] =tan−1tan(π4−x2)=π4−x2.