If tan-1(1+x2-1-x2)/1+x2+1-x2)=alpha ,then prove that x2=sin2alpha

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Q. If

{tan}^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = α

, then

x^{2}

is equal to :

Prove that $t a n^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) = \frac{π}{4} + \frac{1}{2} c o s^{- 1} x^{2} .$

Q. If

\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}})

= α, then x² =
(a) sin 2 α
(b) sin α
(c) cos 2 α
(d) cos α

Q. If

{tan}^{- 1} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}} = α

, then

x^{2} =

Q. If

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

each is equal to

2 {tan}^{- 1} x .

,then show that

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

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Differentiation Using Substitution

Standard XII Mathematics