Prove that ✓1+sinA/1-sinA= secA+tanA​

Step-by-step explanation:

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ANSWER:

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LHS = √(1 + sin∅)/√(1 - sin∅)

= √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅)

= √(1 + sin∅)²/√(1 -sin²∅)

= (1 + sin∅)/√cos²∅

= (1 + sin∅)/cos∅

= 1/cos∅ + sin∅/cos∅

= sec∅ + tan∅ = RHS

Step-by-step explanation:

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Step-by-step explanation:

√1+sinA/1-sinA

=√(1+sinA)²/1-sin²A [rationalizing the denominator]

=√(1+sinA)²/cos²A

=1+sinA/cosA

=secA+tanA

Step-by-step explanation:

√1-sinA/1+sinA=secA-tanA

L.H.S

√1-sinA/1+sinA

multiple by 1-SinA

√1-sinA/1+sinA×1-SinA/1-SinA

√(1-SinA)^2/√1-(Sin^2A)

1-SinA/√Cos^2

1-SinA/CosA

1/CosA-SinA/CosA

SecA-TanA

Hence Proved

✯ The question will be To prove :

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Taking LHS that is

Multiple and divide Nr and Dr with

1+ sin a because it becomes 1 and there will be no affect on the fraction after multiplying by 1

And we know that1- sin²a = cos²a .So ,

Now writing 1 and sin²a separately.

We know that

So replacing these values we got

LHS = ( seca + tan a )² = RHS

hence proved.