Prove that $\tan^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)=\dfrac{\pi}{4}-\dfrac{x}{2},\ x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.

CBSE Class 12 · 2012 · Inverse Trigonometric FunctionsLong Answer

Prove that

tan^{- 1} (\frac{cos x}{1 + sin x}) = \frac{π}{4} - \frac{x}{2}, x \in (- \frac{π}{2}, \frac{π}{2})

Answer: Proved.

Write the inside in half-angle form:

cos x = cos^{2} \frac{x}{2} - sin^{2} \frac{x}{2}

and

1 + sin x = (cos \frac{x}{2} + sin \frac{x}{2})^{2}

\frac{cos x}{1 + sin x} = \frac{( cos \frac{x}{2} - sin \frac{x}{2} ) ( cos \frac{x}{2} + sin \frac{x}{2} )}{( cos \frac{x}{2} + sin \frac{x}{2} ) ^{2}} = \frac{cos \frac{x}{2} - sin \frac{x}{2}}{cos \frac{x}{2} + sin \frac{x}{2}} = \frac{1 - tan \frac{x}{2}}{1 + tan \frac{x}{2}} = tan (\frac{π}{4} - \frac{x}{2})

For

x \in (- \frac{π}{2}, \frac{π}{2})

\frac{π}{4} - \frac{x}{2} \in (0, \frac{π}{2}) \subset (- \frac{π}{2}, \frac{π}{2})

, so

tan^{- 1} (\frac{cos x}{1 + sin x}) = \frac{π}{4} - \frac{x}{2}

\frac{π}{4} - \frac{x}{2}

CBSE Class 12 Board · 2012 · Subjective