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CBSE Class 12 · 2005 · Inverse Trigonometric FunctionsVery Short Answer

Write the following function in the simplest form:

tan^{- 1} \frac{1 - cos x}{1 + cos x}, x < π

.

Answer: $\frac{x}{2}$

Solution

Use the half-angle forms

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

and

1 + cos x = 2 cos^{2} \frac{x}{2}

.

\frac{1 - cos x}{1 + cos x} = \frac{2 sin ^{2} \frac{x}{2}}{2 cos ^{2} \frac{x}{2}} = tan \frac{x}{2}

For

0 < x < π

we have

\frac{x}{2} \in (0, \frac{π}{2})

, so

tan \frac{x}{2} > 0

and the modulus drops.

tan^{- 1} (tan \frac{x}{2}) = \frac{x}{2} (\frac{x}{2} \in (0, \frac{π}{2}))

\frac{x}{2}

CBSE Class 12 Board · 2005 · Subjective

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