Question

Proved that.
√(1 + sinA ÷ 1 - sinA) = secA + tanA.

No copyright answer.​

Answer 1

=1+sinA/cosA

= 1/cosA + sinA /cosA

=secA+tanA

=RHS

Step-by-step explanation:

hope this helps you..

Answer 2

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♧Question♧

$\Huge{ \dashrightarrow {\sf\sqrt{ \frac{1 + sinA}{1 - sinA} } }}$

♧Answer♧

$\large{ \dashrightarrow {\sf\sqrt{ \frac{1 + sinA}{1 - sinA} } }} \\ \\ \large{ \sf{By \: conjugate \: multiplication \: method }} \\ \\ \large{ \dashrightarrow {\sf\sqrt{ \frac{1 + sinA}{1 - sinA} \times{\sf{ \frac{1 + sinA}{1 + sinA}}}} }} \\ \\ \large{ \dashrightarrow {\sf\sqrt{ \frac{({1 + sinA})^{2} }{1 - sin^{2} A} } }} \\ \\ \large{ \dashrightarrow {\sf\sqrt{ \frac{(1 + sinA)^{2} }{cos ^{2} A} } }} \\ \\ \large{ \dashrightarrow {\sf\sqrt{ (\frac{1 + sinA}{cosA} })^{2} }} \\ \\ \large{ \dashrightarrow {\sf{\frac{1 + sinA}{cosA} } }} \\ \\ \large{ \dashrightarrow {\sf{\frac{1}{cosA} + \frac{sinA}{cosA} } }} \\ \\ \sf{we \: know \: that \: \frac{1}{cosA} = secA \: \: and \: \frac{sinA}{cosA} = tanA} \\ \\ \dashrightarrow{ {\boxed{ \sf \red{secA + tanA}}}}$

So proved that,

${\large{ \dashrightarrow {\sf\sqrt{ \frac{1 + sinA}{1 - sinA} } }} = \sf{secA + tanA}}$

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