Question

prove that :- √1+sinA/1-sinA + √1 - sinA/1+sinA = 2 secA

Answer 1

Step-by-step explanation:

Hope it will help you☆☆☆

Answer 2

Step-by-step explanation:

$\bf\underline{\underline{\red{To\:Prove:-}}}$

• $\rm \sqrt{ \dfrac{1 + sinA}{1 - sinA} } + \sqrt{ \dfrac{1 - sinA}{1 + sinA} } = 2secA$

$\bf\underline{\underline{\green{Proof:-}}}$

Let us prove by simplifying LHS and RHS seperately.

$\rm LHS = \sqrt{ \dfrac{1 + sinA}{1 - sinA} } + \sqrt{ \dfrac{1 - sinA}{1 + sinA} }$

By rationalising the denominator:-

$\rm = \sqrt{ \dfrac{1 + sinA}{1 - sinA} \times \dfrac{1 + sinA}{1 + sinA}} + \sqrt{ \dfrac{1 - sinA}{1 + sinA} \times \dfrac{1 - sinA}{1 - sinA} }$

$\rm = \sqrt{ \dfrac{(1 + sinA)(1 + sinA)}{(1 - sinA)(1 + sinA)} }+ \sqrt{ \dfrac{(1 - sinA)(1 - sinA)}{(1 + sinA)(1 - sinA)} }$

$\rm = \sqrt{ \dfrac{(1 + sinA) ^{2} }{(1 )^{2} -( sinA)^{2} } }+ \sqrt{ \dfrac{(1 - sinA)^{2} }{(1 )^{2} -( sinA)^{2}} } \: \: \: \: \: \: \: \: \bigg( \because(a + b)(a - b) = {a}^{2} - {b}^{2} \bigg)$

$\rm = \sqrt{ \dfrac{(1 + sinA) ^{2} }{1 - sin^{2} A} }+ \sqrt{ \dfrac{(1 - sinA)^{2} }{1 - sin ^{2} A} }$

$\rm = \sqrt{ \dfrac{(1 + sinA) ^{2} }{cos ^{2} A} }+ \sqrt{ \dfrac{(1 - sinA)^{2} }{cos ^{2} A} } \: \: \: \: \: \: \: \: \bigg( \because \: 1 - sin^{2} A = {cos}^{2} A \bigg)$

$\rm = \dfrac{ \sqrt{ (1 + sinA) ^{2} }}{ \sqrt{ cos ^{2} A} } + \dfrac{ \sqrt{ (1 - sinA)^{2}} }{ \sqrt{cos ^{2} A}} \:$

$\rm = \dfrac{ 1 + sinA }{ \cos A}+ \dfrac{ 1 - sinA}{ cos A}$

$\rm = \dfrac{ 1 + sinA + 1 - sinA}{ \cos A}$

$\rm = \dfrac{ 2}{ \cos A}$

$\rm = \dfrac{ 2}{ cosA} \: \: \: \: \: \: \: \: \: \: \: \bigg( \because \dfrac{1}{cosA} = secA \bigg)$

$\rm = 2secA= RHS$

LHS = RHS

Hence Proved