Math, asked by hollyphoenix3107, 10 months ago

prove the above question plz.... ​

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Answered by hukam0685
4

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.

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