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CBSE Class 12 · 2015 · Inverse Trigonometric FunctionsLong Answer

If

(tan^{- 1} x)^{2} + (cot^{- 1} x)^{2} = \frac{5 π ^{2}}{8}

, then find

x

.

Answer: $x = - 1$

Solution

Let

t = tan^{- 1} x

; then

cot^{- 1} x = \frac{π}{2} - t

. Substitute:

t^{2} + (\frac{π}{2} - t)^{2} = \frac{5 π ^{2}}{8} \Rightarrow 2 t^{2} - π t + \frac{π ^{2}}{4} = \frac{5 π ^{2}}{8}

2 t^{2} - π t - \frac{3 π ^{2}}{8} = 0 \Rightarrow 16 t^{2} - 8 π t - 3 π^{2} = 0

t = \frac{8 π \pm 64 π ^{2} + 192 π ^{2}}{32} = \frac{8 π \pm 16 π}{32} \Rightarrow t = \frac{3 π}{4} or t = - \frac{π}{4}

Since

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

, only

t = - \frac{π}{4}

is valid, giving

x = tan (- \frac{π}{4}) = - 1

.

x = - 1

CBSE Class 12 Board · 2015 · Subjective

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