# If tantheta + sectheta = x, show that sintheta=(x^2-1)/(x^2+1) ?

Jul 14, 2018

Given, $\sec \theta + \tan \theta = x \ldots . \left[1\right]$

$\rightarrow {\sec}^{2} \theta - {\tan}^{2} \theta = 1$

$\rightarrow \left(\sec \theta + \tan \theta\right) \left(\sec \theta - \tan \theta\right) = 1$

$\rightarrow x \cdot \left(\sec \theta - \tan \theta\right) = 1$

$\rightarrow \sec \theta - \tan \theta = \frac{1}{x} \ldots . . \left[2\right]$

Adding $\left[1\right] \mathmr{and} \left[2\right]$,

$\rightarrow \sec \theta \cancel{+ \tan \theta} + \sec \theta \cancel{- \tan \theta} = x + \frac{1}{x}$

$\rightarrow 2 \sec \theta = \frac{{x}^{2} + 1}{x} \ldots \left[3\right]$

Subtracting $\left[2\right]$ from $\left[1\right]$,we get

$\rightarrow \sec \theta + \tan \theta - \left(\sec \theta - \tan \theta\right) = x - \frac{1}{x}$

$\rightarrow \cancel{\sec \theta} + \tan \theta \cancel{- \sec \theta} + \tan \theta = x - \frac{1}{x}$

$\rightarrow 2 \tan \theta = \frac{{x}^{2} - 1}{x} \ldots . \left[4\right]$

Dividing $\left[4\right]$ by $\left[3\right]$, we get,

$\rightarrow \frac{2 \tan \theta}{2 \sec \theta} = \frac{\frac{{x}^{2} - 1}{x}}{\frac{{x}^{2} + 1}{x}}$

$\rightarrow \sin \theta = \frac{{x}^{2} - 1}{{x}^{2} + 1}$ Proved