Question

[√1-SinA÷√1+sinA] is equals to​

Answer 1

Solution :-

We need to find ,

• $\sf\dfrac{\sqrt{1-sinA}}{\sqrt{1+sinA}}$

It can be written as,

$\implies\sf\sqrt{\dfrac{1-sinA}{1+sinA}}$

By Rationalizing the denominator,

$\implies\sf\sqrt{\dfrac{1-sinA}{1+sinA}\times\dfrac{1-sinA}{1-sinA}}$

$\implies\sf\sqrt{\dfrac{(1-sinA)(1-sinA)}{(1+sinA)(1-sinA)}}$

Using ,

• ( a - b ) ( a - b ) = ( a - b )²

• ( a + b ) ( a - b ) = a² - b²

$\implies\sf\sqrt{\dfrac{(1-sinA)^2}{1^2-sin^2A}}$

$\implies\sf\dfrac{\sqrt{(1-sinA)^2}}{\sqrt{1-sin^2A}}\qquad\;\;\;\Big(\because \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\Big)$

$\implies\sf\dfrac{1-sinA}{\sqrt{cos^2A}}\qquad\;\;\;\Big[\because 1-sin^2A=cos^2A\Big]$

$\sf\implies \dfrac{1-sinA}{cosA}$

$\sf\implies\dfrac{1}{cosA}-\dfrac{sinA}{cosA}$

Substituting ,

• $\sf\dfrac{1}{cosA}$=secA

• $\sf\dfrac{sinA}{cosA}$=tanA

$\implies \underline{\boxed{\dfrac{\sqrt{\textbf{\textsf{1-sinA}}}}{\sqrt{\textbf{\textsf{1+sinA}}}}=\textbf{\textsf{secA-tanA}}}}$