Question

Prove that :

i). 1 + cos A/sin A = sin A/1-cos A​

Answer 1

$\huge\red{\boxed{\bold{Solution}}}$

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

_________________________

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

.

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

.

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

.

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

.

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

hence, LHS = RHS

$\huge{\boxed{\bold{Hence\:proved}}}$

$\huge\mathbb\blue{THANK\:YUH!}$

Answer 2

$\frac{(1+cosA)}{sinA} \\ </p><p> \\ multiply \: by \: sina \: in \: numerator \: and \: denominator \\ we \: get \\ \\ \frac{(1+cosa)sina}{sin^{2} a} \\ \\ \\ now \: we \: know \: that \\ sin^{2}a = (1 - cos^{2}a) \\ and \\ ( {a}^{2} - {b}^{2} ) = (a + b)(a - b) \\ \\ so \: we \: get \\ \frac{(1 + cosa)sina}{(1 + cosa)(1 - cosa)} \\ \\ = \frac{sina}{(1 - cosa)} \\ \\ proved$

$\huge\blue{THANKS}$