# Question #2144a

Dec 18, 2017

It is proved below:

#### Explanation:

$\tan a + \sec a = \frac{x}{y}$
$\mathmr{and} , \sin \frac{a}{\cos} a + \frac{1}{\cos} a = \frac{x}{y}$
$\mathmr{and} , \frac{\sin a + 1}{\cos} a = \frac{x}{y}$
$\mathmr{and} , {\left(\sin a + 1\right)}^{2} / {\cos}^{2} a = {x}^{2} / {y}^{2}$
$\mathmr{and} , {\left(\sin a + 1\right)}^{2} / \left(1 - {\sin}^{2} a\right) = {x}^{2} / {y}^{2}$
$\mathmr{and} , {\left(\sin a + 1\right)}^{2} / \left(\left(1 + \sin a\right) \left(1 - \sin a\right)\right) = {x}^{2} / {y}^{2}$
$\mathmr{and} , \frac{\sin a + 1}{1 - \sin a} = {x}^{2} / {y}^{2}$
$\mathmr{and} , \frac{\sin a + 1 - 1 + \sin a}{\sin a + 1 + 1 - \sin a} = \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} \text{ }$[By division-addition method]
$\mathmr{and} , \frac{2 \sin a}{2} = \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}$
$s o , \sin a = \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} \textcolor{b l u e}{\left[P r o v e d .\right]}$

Dec 18, 2017

Given $\sec \left(a\right) + \tan \left(a\right) = \frac{x}{y} \ldots \ldots \left(1\right)$

We know

${\sec}^{2} \left(a\right) - {\tan}^{2} \left(a\right) = 1$

$\implies \left(\sec \left(a\right) + \tan \left(a\right)\right) \left(\sec \left(a\right) - \tan \left(a\right)\right) = 1$

$\implies \sec \left(a\right) - \sec \left(a\right) = \frac{1}{\sec \left(a\right) + \tan \left(a\right)}$

$\implies \sec \left(a\right) - \tan \left(a\right) = \frac{1}{\frac{x}{y}}$

$\implies \sec \left(a\right) - \tan \left(a\right) = \frac{y}{x} \ldots \ldots . \left(2\right)$

Adding (1) and (2) we get

$\implies 2 \sec \left(a\right) = \frac{x}{y} + \frac{y}{x} = \frac{{x}^{2} + {y}^{2}}{x y}$

$\implies \sec \left(a\right) = \frac{{x}^{2} + {y}^{2}}{2 x y} \ldots . \left(3\right)$

Now subtracting (2) from (1) we get

$\implies 2 \tan \left(a\right) = \frac{x}{y} - \frac{y}{x} = \frac{{x}^{2} - {y}^{2}}{x y}$

$\implies \tan \left(a\right) = \frac{{x}^{2} - {y}^{2}}{2 x y} \ldots . . \left(4\right)$

Dividing (4) by (3) we have

$\tan \frac{a}{\sec} \left(a\right) = \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}$

$\implies \sin \left(a\right) = \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}}$