Prove that $\tan\left(\dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}\dfrac{a}{b}\right)+\tan\left(\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}\dfrac{a}{b}\right)=\dfrac{2b}{a}$.

Proved: the sum equals $\dfrac{2b}{a}$.

CBSE Class 12 · Inverse Trigonometric FunctionsLong Answer

Prove that

tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} \frac{a}{b}) + tan (\frac{π}{4} - \frac{1}{2} cos^{- 1} \frac{a}{b}) = \frac{2 b}{a}

Answer: Proved: the sum equals $\frac{2 b}{a}$ .

Let

θ = \frac{1}{2} cos^{- 1} \frac{a}{b}

, so

cos 2 θ = \frac{a}{b}

. Expand each tangent:

tan (\frac{π}{4} + θ) = \frac{1 + tan θ}{1 - tan θ}, tan (\frac{π}{4} - θ) = \frac{1 - tan θ}{1 + tan θ}

Add:

\frac{1 + tan θ}{1 - tan θ} + \frac{1 - tan θ}{1 + tan θ} = \frac{( 1 + tan θ ) ^{2} + ( 1 - tan θ ) ^{2}}{1 - tan ^{2} θ} = \frac{2 ( 1 + tan ^{2} θ )}{1 - tan ^{2} θ}

Divide numerator and denominator by

cos^{2} θ \to

multiply by

\frac{cos ^{2} θ}{cos ^{2} θ}

\frac{2 ( 1 + tan ^{2} θ )}{1 - tan ^{2} θ} = \frac{2 ( cos ^{2} θ + sin ^{2} θ )}{cos ^{2} θ - sin ^{2} θ} = \frac{2}{cos 2 θ} = \frac{2}{a / b} = \frac{2 b}{a}

\frac{2 b}{a}

CBSE Class 12 Board · Subjective