Question

√1+sinA/1-sinA=secA+tanA​

Answer 1

Step-by-step explanation:

Given :

$\bold{\sqrt{\dfrac{1+sinA}{1-sinA} } }$

To prove :

$\bold{\sqrt{\dfrac{1+sinA}{1-sinA} } =secA+ tan A}$

Solution :

• Rationalizing the denominator.

$\rightarrowtail \bold{\sqrt{\dfrac{1+sinA}{1-sinA} }\times \sqrt{\dfrac{1+sinA}{1+sinA} } }$

$\rightarrowtail \bold{\sqrt{\dfrac{(1+sinA)^{2}}{(1-sin)^{2}} } }$

$\rightarrowtail \bold{\dfrac{1+sinA}{1-sin^{2}} }$

• ∵ sin²A + cos²A = 1
• ∴ 1 - cos²A = sin²A

$\rightarrowtail \bold{\dfrac{1-sinA}{cosA} }$

$\mapsto \bold{\dfrac{1}{cosA}+\dfrac{sinA}{cosA} }$

• ∵ 1/cosA = secA
• sinA/cosA = tanA

$\rightarrowtail \bold{secA+ tanA}$

• ∴ L.H.S = R.H.S