Question

(1-sinA)/(1+sinA)=(secA-tanA)^2​

Answer 1

Answer:

step by step correct explanation

Answer 2

HEY MATE YOUR ANSWER IS HERE....

TAKING L.H.S

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

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

$\frac{1 + { \sin }^{2}a \: - 2 \sin(a) }{1 - { \sin }^{2}a }$

$\frac{1}{ { \cos }^{2}a } + \frac{ { \sin }^{2}a }{ { \cos }^{2}a } - \frac{2 \sin(a) }{ { \cos }^{2}a }$

${ \sec }^{2} a + \ { \tan }^{2} a - 2( \frac{ \sin(a) }{ \cos(a) \: } \times \frac{1}{ \cos(a) })$

${ \sec }^{2} a + { \tan}^{2} a \: - 2 \tan(a) \sec(a)$

$({ \sec \: a \: - \tan \: a })^{2}$

HENCE PROVED

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