Question

cosA/1+sinA+tanA=secA

Answer 1

Given : cosA/1 + sinA + tanA

$= \ \textgreater \ \frac{cosA}{1 + sinA} + \frac{sinA}{cosA}$

$= \ \textgreater \ \frac{cos^2A + sinA(1 + sinA)}{(1 + sinA)(cosA)}$

$= \ \textgreater \ \frac{cos^2A + sinA + sin^2A}{(1 + sinA)(cosA)}$

$= \ \textgreater \ \frac{1 + sinA}{(1 + sinA)(cosA)}$

$= \ \textgreater \ \frac{1}{cosA}$

SecA.


Hope this helps!

Answer 2

Hey !!!

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 prooved here ✒

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Hope it helps you !!

@Rajukumar111