Math, asked by Revanth2299, 11 months ago

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

Answers

Answered by MBsquad
0

Step-by-step explanation:

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Answered by MaIeficent
7

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

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