Question

if sin thete = x-y/x+y, then tan(pi/4+theta/2) may be equal to​

Answer 1

Step-by-step explanation:

Given,

$\sin( \alpha ) = \frac{x - y}{x + y}$

Now,

$\tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) = \frac{ \tan( \frac{\pi}{4} ) + \tan( \frac{ \alpha }{2} ) }{1 - \tan( \frac{\pi}{4} ) \tan( \frac{ \alpha }{2} ) }$

$= \tan( \frac{ \pi}{4} + \frac{ \alpha }{2} ) = \frac{1 + \tan (\frac{ \alpha }{2}) }{1 - \tan( \frac{ \alpha }{2} ) }$

Now, break tan into sin and cos and take lcm

$\tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) = \frac{ \cos( \frac{ \alpha }{2} ) + \sin( \frac{ \alpha }{2} ) }{ \cos( \frac{ \alpha }{2} ) - \sin( \frac{ \alpha }{2} ) }$

Multiplyng ( cos α+ sin α) in numerator and denominator, we get,

$\tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) = \frac{1}{ { (\cos( \frac{ \alpha }{2} ) })^{2} - ( { \sin( \frac{ \alpha }{2} ) })^{2} }$

$\tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) = \frac{1}{ \cos( \alpha ) }$

As, (cos² θ-sin²θ)=cos 2θ

Now,

$\tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) = \frac{1}{ \sqrt{1 - ( \sin( \alpha ))^{2} } } = \frac{1}{ \sqrt{1 - \frac{(x - y) ^{2} }{ ({x + y})^{2} } } }$

Taking lcm and symplifying, we get,

$\tan( \frac{\pi}{4} + \frac{ \alpha }{2} ) = \frac{x + y}{2 \sqrt{xy} }$

Hope, this help you!