Question

√(1- Cos A) /
√(1 + cos A) =
sin A/
1+ COS A​

Answer 1

Solution:

$\sqrt{ \dfrac{ 1 - Cos\:A}{1 + Cos\:A}} = \dfrac{Sin\:A}{1 + Cos\:A}\\\\\\\text{Taking RHS we get,}\\\\\text{According to Trigonometric Identity}\\\\\boxed{ Sin\:A = \sqrt{ 1 - Cos^2\:A}}\\\\\\\implies RHS = \dfrac{ \sqrt{ 1 - Cos^2\:A}}{1 + Cos\:A}}\\\\\text{Taking the denominator inside the square root, we get:}\\\\\implies RHS = \sqrt{ \dfrac{ 1 - Cos^2\:A}{(1 + Cos\:A)^2}}$

$\text{Since the numerator is of the form}\:\: a^2 - b^2, \text{we get}\\\\\\\implies RHS = \sqrt{ \dfrac{( 1 + Cos\:A)(1 - Cos\:A)}{(1 + Cos\:A)(1 + Cos\:A)}}\\\\\\\text{ ( 1 + Cos A ) gets cancelled. Hence we get:}\\\\\\\implies RHS = \sqrt{ \dfrac{1 - Cos\:A}{1 + Cos\:A }} = LHS\\\\\\\textbf{Hence Proved}$