Question

prove that,secA-tanA/secA+tanA=cos^2A/(1+sinA)^2​

Answer 1

$\frac{ \sec A - \tan A }{ \sec A + \tan A } = \frac{ { \cos}^{2} A}{ {(1 + \sin A) }^{2} }$

L.H.S,

$\frac{ \sec A - \tan A }{ \sec A + \tan A }$

$= \frac{ \frac{1}{ \cos A } - \frac{ \sin A }{ \cos A } }{\frac{1}{ \cos A } + \frac{ \sin A }{ \cos A }}$

$= \frac{ \frac{1 - \sin A }{ \cos A } }{\frac{1 + \sin A }{ \cos A} }$

$= {\frac{1 - \sin A }{ \cos A}} \times \frac{ \cos A }{1 + \sin A }$

$= \frac{ 1 - \sin A }{1 + \sin A}$

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

$= \frac{1 - { \sin }^{2} A}{ {(1 + \sin A )}^{2} }$

$= \frac{ { \cos}^{2} A}{ {(1 + \sin A) }^{2} }$

⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀[ as 1-sin²A = cos²A]

= R.H.S

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀[hence proved]