cosA/1+sinA+tanA=secA
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1
Given : cosA/1 + sinA + tanA
= \ \textgreater \ \frac{cosA}{1 + sinA} + \frac{sinA}{cosA}= >
1+sinA
cosA
+
cosA
sinA
= \ \textgreater \ \frac{cos^2A + sinA(1 + sinA)}{(1 + sinA)(cosA)}= >
(1+sinA)(cosA)
cos
2
A+sinA(1+sinA)
= \ \textgreater \ \frac{cos^2A + sinA + sin^2A}{(1 + sinA)(cosA)}= >
(1+sinA)(cosA)
cos
2
A+sinA+sin
2
A
= \ \textgreater \ \frac{1 + sinA}{(1 + sinA)(cosA)}= >
(1+sinA)(cosA)
1+sinA
= \ \textgreater \ \frac{1}{cosA}= >
cosA
1
SecA.
Hope this helps!
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