Solve $\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\dfrac{8}{31}$.

CBSE Class 12 · Inverse Trigonometric FunctionsLong Answer

Solve

tan^{- 1} (x + 1) + tan^{- 1} (x - 1) = tan^{- 1} \frac{8}{31}

Answer: $x = \frac{1}{4}$

Combine the left side:

tan^{- 1} [\frac{( x + 1 ) + ( x - 1 )}{1 - ( x + 1 ) ( x - 1 )}] = tan^{- 1} \frac{8}{31} \Rightarrow \frac{2 x}{1 - ( x ^{2} - 1 )} = \frac{8}{31}

\frac{2 x}{2 - x ^{2}} = \frac{8}{31} \Rightarrow 62 x = 16 - 8 x^{2} \Rightarrow 8 x^{2} + 62 x - 16 = 0 \Rightarrow 4 x^{2} + 31 x - 8 = 0

(4 x - 1) (x + 8) = 0 \Rightarrow x = \frac{1}{4} or x = - 8

For

x = - 8

the left side is a sum of two negative angles (negative), while the right side is positive; reject it.

x = \frac{1}{4}

(Source note: the stem was printed as

tan^{- 1} \frac{18}{31}

; the intended value is

\frac{8}{31}

, confirmed by the worked key and by the duplicate Q29.)

CBSE Class 12 Board · Subjective