Question

Prove that
$\sqrt{1 + sina) \div \sqrt{1 - sina}$=secA+tanA
​

Answer 1

Questions

$\tt \to \: \dfrac{ \sqrt{1 + sinA} }{ \sqrt{1 - sinA} } = secA + tanA$

Solution

Now Take LHS

$\tt \to \: \dfrac{ \sqrt{1 + sinA} }{ \sqrt{1 - sinA} }$

We can write as

$\tt \to \: \sqrt{\dfrac{1 + sinA}{1 - sinA } }$

Using Rationalization Method

$\tt \to \: \sqrt{\dfrac{1 + sinA}{1 - sinA } \times \dfrac{1 + sinA}{1 + sinA} }$

Using this identities

$\tt \to \: (a + b)(a + b) = (a + b)^{2}$

$\tt \to \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

we get

$\tt \to \: \sqrt{\dfrac{(1 + sinA) {}^{2} }{1 - sin^{2} A } }$

We Know that

$\tt \to \:1 - sin^{2} \theta = cos^{2} \theta$

we get

$\tt \to \: \sqrt{\dfrac{(1 + sinA) {}^{2} }{cos {}^{2} A } }$

$\tt \to \: \dfrac{1 + sinA}{cosA} = \dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

$\tt \to \: secA + tanA$

Hence Proved