Write in the simplest form: $\tan^{-1}\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$, where $0<x<\dfrac{\pi}{2}$.

CBSE Class 12 · 2007 · Inverse Trigonometric FunctionsVery Short Answer

Write in the simplest form:

tan^{- 1} (\frac{1 + sin x + 1 - sin x}{1 + sin x - 1 - sin x})

, where

0 < x < \frac{π}{2}

Answer: $\frac{π}{2} - \frac{x}{2}$

Write

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

. For

0 < x < \frac{π}{2}

cos \frac{x}{2} > sin \frac{x}{2} > 0

, so

1 + sin x = cos \frac{x}{2} + sin \frac{x}{2}, 1 - sin x = cos \frac{x}{2} - sin \frac{x}{2} .

Hence

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

Since

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

tan^{- 1} (tan (\frac{π}{2} - \frac{x}{2})) = \frac{π}{2} - \frac{x}{2}

\frac{π}{2} - \frac{x}{2}

(Note: the

cot^{- 1}

of this same expression equals

\frac{x}{2}

; for

tan^{- 1}

the answer is its complement.)

CBSE Class 12 Board · 2007 · Subjective