Question

Prove that:-

(1-sinx)/(1+sinx)=(secx-tanx)^2.

Answer 1

$\huge \bf{HEY \: FRIENDS!!}$


----------------------------------------------------



$\huge \bf \underline{Here \: is \: your \: answer↓}$


⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇



$\huge \boxed{Prove \: \: that:-)}$


$\bf = > \frac{1 - \sin(x) }{1 + \sin(x) } = {( \sec(x) - \tan(x) )}^{2} .$


$\huge \boxed{Solving \: RHS.}$


$\bf = {( \sec(x) - \tan(x) ) }^{2} .$


$\bf = {( \frac{1}{ \cos(x) } - \frac{ \sin(x) }{ \cos(x) }) }^{2}.$


$\bf = {( \frac{1 - \sin(x) }{ \cos(x) } )}^{2} .$


$\bf = \frac{ {(1 - \sin(x) )}^{2} }{ { \cos}^{2}(x)} .$


$\bf = \frac{ {(1 - \sin(x)) }^{2} }{(1 - { \sin }^{2} (x)) } .$


$\bf = \frac{(1 - \sin(x))(1 - \sin(x)) }{(1 + \sin(x) )(1 - \sin(x) )} .$


$\huge \bf = \frac{1 - \sin(x) }{1 + \sin(x) } .$


$\huge \bf \underline{RHS = LHS.}$


✅✅ Hence, it is proved ✔✔.



$\huge \boxed{THANKS}$


$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: you}$

Answer 2

$\huge \bf {hey}$
$\huge \underline{here \: is \: the \: answer}$
⬇⤵⏬

$\bf{ \frac{1 - sin(x)}{1 + sin(x)} = ( {secx - tanx}^{2}) }$
$\huge{solving \: rhs}$

$\bf{( {secx - tanx)}^{2} }$

▶
${ \frac{1}{cosx} }^{2} - ({ \frac{sinx}{cosx} }^{2} )$
▶

(1-sinx)²/1-sin²x

$\bf{ \frac{(1 - sinx)(1 - sinx)}{(1 + sinx)(1 - sinx)} }$

▶

$\bf{ \frac{1 - sinx}{1 + sinx} }$


$\huge \boxed{rhs = lhs}$



$\huge{hence \: it \: is \: proved}$
$\huge \underline{hope \:it \: helps}$
$\huge{thanks}$