Prove that (cosA/1+sinA)+tanA=secA
Answers
Answered by
0
Answer: Please mark my answer the brainliest!
from LHS
cosA /1 + sinA + tana
CosA /1 + sinA × 1 - sinA /1 - sinA + tanA
[here , by ( 1 - sinA) multiplied on numerator and denominator cosA/1 + sinA ] or in simple way rasionalising process of (cosA/1 + sinA )
we get ,
cos ( 1 - sinA )
-------------------------- + tanA
( 1 + sinA ) ( 1 - sinA )
=> cos ( 1 - sinA )
----------------------- + tanA
cos²A
=> 1 - sinA / cosA + tanA
=> secA - tanA + tanA
=> secA RHS
Similar questions