Question

Can someone pls help..... (Class XI Trigonometry)

Answer 1

Step-by-step explanation:

Given,

$\tan^{2} ( \frac{\pi}{4} + \frac{ \theta}{2} ) = \frac{a}{b}$

$= > \frac{2 \sin^{2} ( \frac{\pi}{4} + \frac{\theta}{2} ) }{ 2\cos^{2} ( \frac{\pi}{4} + \frac{\theta}{2} ) } = \frac{a}{b}$

$= > \frac{ 1 - \cos2( \frac{\pi}{4} + \frac{ \theta }{2} ) }{1 + \cos2( \frac{\pi}{4} + \frac{\theta}{2} ) } = \frac{a}{b}$

$= > \frac{1 - \cos( \frac{\pi}{2} + \theta) }{1 + \cos( \frac{\pi}{2} + \theta)} = \frac{a}{b}$

$= > \frac{1 + \sin( \theta ) }{1 - \sin( \theta) } = \frac{a}{b}$

Now, using componendo and dividendo,

$\frac{1 + \sin( \theta) + 1 - \sin( \theta) }{1 + \sin( \theta) - 1 + \sin( \theta)} = \frac{a + b}{a - b}$

$= > \frac{2}{2 \sin( \theta ) } = \frac{a + b}{a - b}$

$= > \frac{1}{ \sin( \theta ) } = \frac{a + b}{a - b}$

$= > \sin( \theta) = \frac{a - b}{a + b}$