Math, asked by dh26, 1 month ago

1+sin A/cos A+cos A/1+sin A=2secA​

Answers

Answered by Tan201
1

Step-by-step explanation:

We need to prove, \frac{1+sinA}{cosA}+\frac{cosA}{1+sinA}=2secA

LHS=\frac{1+sinA}{cosA}+\frac{cosA}{1+sinA}

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

\frac{1^2+sin^2A+2sinA}{cosA(1+sinA)} +\frac{cos^2A}{cosA(1+sinA)} ((a+b)^2=a^2+b^2+2ab)

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

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

\frac{2+2sinA}{cosA(1+sinA)}

\frac{2(1+sinA)}{cosA(1+sinA)}

\frac{2}{cosA}

2secA=RHS (\frac{1}{cosA} =secA)

LHS=RHS

\frac{1+sinA}{cosA}+\frac{cosA}{1+sinA}=2secA

∴ Hence proved.

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