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CBSE Class 12 · 2008 · Inverse Trigonometric FunctionsLong Answer

Solve for

x

:

tan^{- 1} (2 x) + tan^{- 1} (3 x) = \frac{π}{4}

.

Answer: $x = \frac{1}{6}$

Solution

Using

tan^{- 1} a + tan^{- 1} b = tan^{- 1} \frac{a + b}{1 - ab}

:

tan^{- 1} \frac{2 x + 3 x}{1 - 6 x ^{2}} = \frac{π}{4} \Rightarrow \frac{5 x}{1 - 6 x ^{2}} = 1

5 x = 1 - 6 x^{2} \Rightarrow 6 x^{2} + 5 x - 1 = 0 \Rightarrow (6 x - 1) (x + 1) = 0

So

x = \frac{1}{6}

or

x = - 1

. For

x = - 1

both

tan^{- 1} (2 x)

and

tan^{- 1} (3 x)

are negative and their sum cannot equal

\frac{π}{4} > 0

(also

1 - 6 x^{2} < 0

), so it is rejected.

x = \frac{1}{6}

CBSE Class 12 Board · 2008 · Subjective

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