Question

prove that whole root 1 + sin A upon 1 minus sin A equal to sec A upon tan A​

Answer 1

Correct Question :--- Prove that √[(1+SinA)/(1-sinA)] = secA + tanA

Solution :---

Rationlizing the Denominator by Multiplying √(1+sinA)/√(1+sinA) in LHS we get,

√(1 + sinA)/√(1 - sinA)

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

Using (a+b)(a-b) = a² - b² in denominator we get,

= √(1 + sinA)²/√(1 -sin²A)

using (1-sin²A = cos²A in denominator now,

= (1 + sinA)/√cos²A

= (1 + sinA)/cosA

= 1/cosA + sinA/cosA

Now, we know that , 1/cosA = secA and sinA/cosA = tanA .

= secA + tanA = RHS

✪✪ Hence Proved ✪✪

Similar Question :--

prove that cosA-sinA+1/cos A+sinA-1=cosecA+cotA

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Answer 2

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