Question

prove the above question plz.... ​

Answer 1

Step-by-step explanation:

To prove the trigonometric expression

$\sqrt{ \frac{1 - cos \: \theta}{1 + cos \: \theta} }$

We know that on multiply numerator and denominator by same expression the fraction can not change,so multiply both by

$1 + cos \theta$

So,

$\sqrt{ \frac{1 - cos \: \theta}{1 + cos \: \theta} \times \frac{1 + cos \: \theta}{1 + cos \: \theta } }$

$\sqrt{ \frac{ {(1)}^{2} - cos ^{2} \: \theta}{(1 + cos \: \theta)^{2}}} \\ \\ \because \: 1 - {cos}^{2} \theta = {sin}^{2} \theta \\ \\ \sqrt{ \frac{ 1 - cos ^{2} \: \theta}{(1 + cos \: \theta)^{2}}}\\\\ \therefore \\ \\ \: \sqrt{ \frac{ sin ^{2} \: \theta}{(1 + cos \: \theta)^{2}}}$

and we know that if numerator and denominator both have whole square terms, under root cancels

$\sqrt{ \bigg({ \frac{sin \theta}{1 + cos \theta} \bigg)}^{2} } \\ \\ = \frac{sin \theta}{1 + cos \theta} \\ \\ = RHS$

Hence proved.