Prove that 12cos-1 1-x-1-x -tan-1-x

Let y = 12(cos^ 11 x1+x ) cos 2y = 1 x1+x 1 tan^2 y1+tan^2y =1 x1+x 1 tan^2y +x xtan^2y = 1+ tan^2 y #160; x xtan^2y 2x= 2ta ...

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Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Prove that ...

Question

Prove that $\frac{1}{2} {cos}^{- 1} (\frac{1 - x}{1 + x}) = {tan}^{- 1} \sqrt{x}$

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Solution

Let $y = \frac{1}{2} ({cos}^{- 1} \frac{1 - x}{1 + x})$
$cos 2 y = \frac{1 - x}{1 + x}$
$\frac{1 - {tan}^{2} y}{1 + {tan}^{2} y}$ $= \frac{1 - x}{1 + x}$
$1 - {tan}^{2} y + x - x {tan}^{2} y = 1 + {tan}^{2} y - x - x {tan}^{2} y$
$2 x = 2 {tan}^{2} y$
$x = {tan}^{2} y$
$y = {tan}^{- 1} \sqrt{x}$

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Prove that 12cos−1(1−x1+x)=tan−1√x

Let y=12(cos−11−x1+x)cos2y=1−x1+x1−tan2y1+tan2y =1−x1+x1−tan2y+x−xtan2y=1+tan2y−x−xtan2y2x=2tan2yx=tan2yy=tan−1√x

Prove that $\frac{1}{2} {cos}^{- 1} (\frac{1 - x}{1 + x}) = {tan}^{- 1} \sqrt{x}$

Let $y = \frac{1}{2} ({cos}^{- 1} \frac{1 - x}{1 + x})$
$cos 2 y = \frac{1 - x}{1 + x}$
$\frac{1 - {tan}^{2} y}{1 + {tan}^{2} y}$ $= \frac{1 - x}{1 + x}$
$1 - {tan}^{2} y + x - x {tan}^{2} y = 1 + {tan}^{2} y - x - x {tan}^{2} y$
$2 x = 2 {tan}^{2} y$
$x = {tan}^{2} y$
$y = {tan}^{- 1} \sqrt{x}$