Question

1/1+sinA + 1/1-sinA = 2secA​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\sf \dfrac{1}{1+sinA}+\dfrac{1}{1-sinA}=2.sec^2A$

$\;\;\underline{\textbf{\textsf{ To prove :-}}}$

• $\sf \dfrac{1}{1+sinA}+\dfrac{1}{1-sinA}=2.sec^2A$

$\;\;\underline{\textbf{\textsf{ Proof :-}}}$

LHS

$\dashrightarrow \sf \dfrac{1}{1+sinA}+\dfrac{1}{1-sinA}\\\\ \dashrightarrow \sf \dfrac{(1-sinA)+(1+sinA)}{(1+sinA)(1-sinA)}\\\\ \dashrightarrow \sf \dfrac{2}{1-sin^2A}\\\\\ \dashrightarrow \sf \dfrac{2}{cos^2A}\\\\\to \sf 2.sec^2A\\\\ \dashrightarrow \sf RHS$

Hence, (proved)

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ㅤ$\;\;\underline{\textbf{\textsf{ Need to know :-}}}$

$\ \; \sf (a+b)(a-b)=a^2-b^2\\\ \; \sf cos^2\theta=1-sin^2\theta\\\ \; \sf sec\theta=\dfrac{1}{cos\theta}$

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Answer 2

L.H.S=1/1+sinA+1/1-sinA

=1-sinA+1+sinA /(1+sinA)(1-sinA)

=2-sinA+sinA /(1-sin^2A)

=2/cos^2A

=2× (1/cos^2A)

=2sec^2A which is equal to R.H.S

so it is prove !

hope it will be help you