Question

answer this question don't give irrelevant answers .
icse book
trignomatry is chapter. ​

Answer 1

Step-by-step explanation:

Given,

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

$= > \frac{ \frac{1}{ \cos( \alpha ) } - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }{ \frac{1}{ \cos( \alpha ) } + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } = \frac{1}{4}$

Taking lcm and simplying, we get,

$= > \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } = \frac{1}{4}$

Using componendo and dividendo, we get,

$= > \frac{1 - \sin( \alpha ) + 1 + \sin( \alpha ) }{1 - \sin( \alpha ) - 1 - \sin( \alpha ) } = \frac{1 - 4}{1 + 4}$

$= > \frac{2}{ - 2 \sin( \alpha ) } = \frac{ - 3}{5}$

$= > \sin( \alpha ) = \frac{5}{3}$

But this is not possible... as the value of sine belongs to [-1,1]