Question

cose
2. Prove that
1+ sin 0
= sec o -tan o.

Answer 1

Step-by-step explanation:

sec o - tan o = 1/cos o - sin o / cos o

by taking lcm u will get 1 - sin o / cos o -------[a]

now multiply [a] by cos o / cos o

u will get cos ^2 o

now put value of cos^2 i.e 1 - sin^2 o

u get

1 - sin o ( cos o) / 1 - sin ^2 o

= 1 - sin o ( cos o) / 1 - sin o ( 1 + sin o ).---[ a^2- b^2]

now after cutting 1 - sin o from up n down u get

cos o / 1 + sin o

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