Question

Find the value of √1+sinA/1-sinA

Answer 1

Answer:

L.H.S.=√1+sinA/1-sinA

=√(1+sinA)(1+sinA)/(1-sinA)(1+sinA)

=√(1+sinA)^/1-sin^A

=√(1+sinA)^/cos^A

=1+sinA/cosA

= 1/cosA+ sinA/cosA

=secA+tanA

=R.H.S. prove

Answer 2

$\displaystyle \sf{ \sqrt{ \frac{1 + sin A}{1 - sin A} } } = sec A + tan A$

Given :

$\displaystyle \sf{ \sqrt{ \frac{1 + sin A}{1 - sin A} } }$

To find :

The value of the expression

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

$\displaystyle \sf{ \sqrt{ \frac{1 + sin A}{1 - sin A} } }$

Step 2 of 2 :

Simplify the given expression

$\displaystyle \sf{ \sqrt{ \frac{1 + sin A}{1 - sin A} } }$

$\displaystyle \sf{ = \sqrt{ \frac{(1 + sin A)(1 + sin A)}{(1 + sin A)(1 - sin A)} } }$

$\displaystyle \sf{ = \sqrt{ \frac{{(1 + sin A)}^{2} }{1 - {sin}^{2} A}}}$

$\displaystyle \sf{ = \sqrt{ \frac{{(1 + sin A)}^{2} }{{cos}^{2} A}}}$

$\displaystyle \sf{ = \frac{{(1 + sin A)}^{} }{{cos}^{} A}}$

$\displaystyle \sf{ = \frac{{1}^{} }{{cos}^{} A} + \frac{{ sin A}^{} }{{cos}^{} A}}$

$\displaystyle \sf{ = sec A+ tan A}$

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