Prove that $\tan^{-1}\left(\dfrac{6x-8x^{3}}{1-12x^{2}}\right)-\tan^{-1}\left(\dfrac{4x}{1-4x^{2}}\right)=\tan^{-1}2x;\ |2x|<\dfrac{1}{\sqrt{3}}$.

CBSE Class 12 · Inverse Trigonometric FunctionsLong Answer

Prove that

tan^{- 1} (\frac{6 x - 8 x ^{3}}{1 - 12 x ^{2}}) - tan^{- 1} (\frac{4 x}{1 - 4 x ^{2}}) = tan^{- 1} 2 x; ∣2 x ∣ < \frac{1}{3}

Answer: Proved.

Let

u = 2 x

. The second argument is

\frac{4 x}{1 - 4 x ^{2}} = \frac{2 u}{1 - u ^{2}}

, and the first is

\frac{6 x - 8 x ^{3}}{1 - 12 x ^{2}} = \frac{3 u - u ^{3}}{1 - 3 u ^{2}}

.
These are the double- and triple-angle forms of the tangent-inverse:

tan^{- 1} \frac{2 u}{1 - u ^{2}} = 2 tan^{- 1} u, tan^{- 1} \frac{3 u - u ^{3}}{1 - 3 u ^{2}} = 3 tan^{- 1} u

(valid since

∣ u ∣ = ∣2 x ∣ < \frac{1}{3}

, keeping the arguments in range). Therefore

tan^{- 1} \frac{6 x - 8 x ^{3}}{1 - 12 x ^{2}} - tan^{- 1} \frac{4 x}{1 - 4 x ^{2}} = 3 tan^{- 1} u - 2 tan^{- 1} u = tan^{- 1} u = tan^{- 1} 2 x

tan^{- 1} 2 x

CBSE Class 12 Board · Subjective