plz help...............
Answers
Step-by-step explanation:
This is your solution:-
To prove:-
\frac{1-sin \theta}{1 +sin \theta} = ( \sec \theta - tan \theta)^{2} \\ \\
Solution:-
L.H.S
\frac{1-sin \theta}{1 +sin \theta} \\ = \frac{(1 - \sin \theta)(1 - \sin \theta) }{(1 + \sin \theta ) (1 - \sin \theta )} \\ = \frac{(1 - \sin \theta)^{2} }{ {(1)}^{2} - (\sin \theta)^{2} \ } \\ = \frac{ {(1 - \sin \theta )}^{2} }{1 - \ { \sin }^{2} \theta } \\ = \frac{ {(1 - \sin \theta )}^{2} }{cos^{2} \theta} \\ = ( \frac{1 - \sin \theta }{cos \theta} )^{2} \\ = (\frac{1}{ \cos \theta } - \frac{ \sin \theta}{ \cos\theta } )^{2} \\ = ( \sec\theta - tan\theta)^{2} \\
Therefore,
\frac{1-sin \theta}{1 +sin \theta} = ( \sec \theta - tan \theta)^{2} \\ \\
L.H.S = R.H.S
Hence proved.
Hope it will help you.
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Answer:
Step-by-step explanation:
