Question

Here is the question

Solve

${tan}^{ - 1} ( \dfrac{sinx}{1 + cosx} )$
​

Answer 1

Step-by-step explanation:

tan^_1 (sin x/1+cos x)

• tan^_1 {2sin(x/2)cos(x/2)/2cos^2(x/2)}

• tan^_1 {sin(x/2)/cos(x/2)}
• tan^_1 {tan(x/2)}
• x/2

Answer 2

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Find the value of :

$\large{ \sf{tan {}^{ - 1} ( \frac{sin \: x}{1 + cos \: x} )}}$

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Formulas to be used

$\sf{sin2 \theta = 2sin \theta \: cos \theta} \\ \\ \sf{2 {cos}^{2} \frac{ \theta}{2} = 1 + cos \theta }$

Now,

$\rightarrow \: \sf{ {tan}^{ - 1} (\frac{2sin \frac{x}{2} .cos \frac{x}{2} }{2 {cos}^{2} \frac{x}{2} } )} \\ \\ \rightarrow \: \sf{ {tan}^{ - 1}( \frac{2 sin \frac{x}{2} }{2cos \frac{x}{2} } ) } \\ \\ \rightarrow \: \sf{ {tan}^{ - 1}( tan \: \frac{x}{2} )} \\ \\ \huge{ \rightarrow \: \tt{ \frac{x}{2} }}$