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CBSE Class 12 · 2024 · Inverse Trigonometric FunctionsShort Answer

If

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

and

b = tan^{- 1} (3) - cot^{- 1} (- \frac{1}{3})

, then find the value of

a + b

.

Answer: $\frac{7 π}{12}$

Solution

Note

\frac{2}{2} = \frac{1}{2}

, so

sin^{- 1} \frac{1}{2} = \frac{π}{4}

and

cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}

.

a = \frac{π}{4} + \frac{2 π}{3} = \frac{11 π}{12}

Also

tan^{- 1} 3 = \frac{π}{3}

and

cot^{- 1} (- \frac{1}{3}) = π - \frac{π}{3} = \frac{2 π}{3}

(angle in

(0, π)

).

b = \frac{π}{3} - \frac{2 π}{3} = - \frac{π}{3}

a + b = \frac{11 π}{12} - \frac{π}{3} = \frac{11 π - 4 π}{12} = \frac{7 π}{12}

\frac{7 π}{12}

CBSE Class 12 Board · 2024 · Subjective

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