Math, asked by vbj16, 6 months ago

cosA /1+ sinA+tanA=secA.


Answers

Answered by jagtarbawa75
2

Step-by-step explanation:

Given : cosA/1 + sinA + tanA

= \ \textgreater \ \frac{cosA}{1 + sinA} + \frac{sinA}{cosA}= \textgreater 1+sinAcosA+cosAsinA

= \ \textgreater \ \frac{cos^2A + sinA(1 + sinA)}{(1 + sinA)(cosA)}= \textgreater (1+sinA)(cosA)cos2A+sinA(1+sinA)

= \ \textgreater \ \frac{cos^2A + sinA + sin^2A}{(1 + sinA)(cosA)}= \textgreater (1+sinA)(cosA)cos2A+sinA+sin2A

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

= \ \textgreater \ \frac{1}{cosA}= \textgreater cosA1

SecA.

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