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CBSE Class 12 · 2011 · Inverse Trigonometric FunctionsLong Answer

Prove that

cot^{- 1} (\frac{1 + sin x + 1 - sin x}{1 + sin x - 1 - sin x}) = \frac{x}{2}, x \in (0, \frac{π}{4})

.

Answer: Proved: expression $= \frac{x}{2}$ .

Solution

Write

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

. For

x \in (0, \frac{π}{4})

,

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}

Then

\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}

Since

\frac{x}{2} \in (0, \frac{π}{8}) \subset (0, π)

,

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

\frac{x}{2}

CBSE Class 12 Board · 2011 · Subjective

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