Question

Prove that tan (√1+x+√1-x) /(√1+x+√1-x) =π/4+1/2cos^-1(x)​

Answer 1

Answer:

To prove

tan-1[1+x-1-x1+x+1-x]=π4-12cos-1x,-12≤x≤1

Taking LHS, we get:

tan-1[1+x-1-x1+x+1-x]

let x=cos2θ

tan-1[1+x-1-x1+cos2θ+1-cos2θ]=tan-1[1+cos2θ-1-cos2θ1+cos2θ+1-cos2θ]

=tan-1[cosθ-sinθcosθ+sinθ]

=tan-1[1-tanθ1+tanθ]

=tan-1tan(π4-θ)

=(π4-θ)

π

θ=π4−θ

π

=π4−12cos−1 x

=RHS

Hence proved.

Answer 2

${tan}^{ - 1} ( \frac{ \sqrt{1 + x} - \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } ) \\ \\ put \: \: \: \: x = cos2 \theta \\ \\ 2 \theta = {cos}^{ - 1} x \\ \\ \theta = \frac{1}{2} {cos}^{ - 1}x \\ \\ {tan}^{ - 1} ( \frac{ \sqrt{1 + cos2 \theta} - \sqrt{1 - cos 2\theta} }{ \sqrt{1 + cos2 \theta} + \sqrt{1 - cos \theta} } ) \\ \\ = {tan}^{ - 1} ( \frac{ \sqrt{2 {cos}^{2} \theta } - \sqrt{ 2{sin}^{2} \theta} }{ \sqrt{2 {cos}^{2} \theta} + \sqrt{2 {sin}^{2} \theta } } ) \\ \\ = {tan}^{ - 1} ( \frac{cos \theta - sin \theta}{cos \theta + sin \theta} ) \\ \\ = {tan}^{ - 1} ( \frac{1 - tan \theta}{1 + tan \theta} ) \\ \\ = {tan}^{ - 1} tan( \frac{\pi}{4} - \theta) \\ \\ = \frac{\pi}{4} - \theta \\ \\ = \frac{\pi}{4} - \frac{1}{2} {cos}^{ - 1} x$