prove that √(1+sinA/1-sinA) + √(1-sinA/1+sinA) = 2secA.​

Question

prove that √(1+sinA/1-sinA) + √(1-sinA/1+sinA) = 2secA.​

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Julia 3 months 2021-11-06T14:38:16+00:00 1 Answer 0 views 0

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    2021-11-06T14:40:08+00:00

    Answer:

    Given:

    The equation is,

    (1+sinA/1sinA)+(1sinA/1+sinA)=2secA

    To prove:

    LHS=RHS

    Solution:

    First take LHS,

    √(1+sinA/1-sinA)+√(1-sinA/1+sinA)

    on rationalizing,

    =(1+sinA/1sinA)×1+sinA/1+sinA)+(1sinA/1+sinA)×1sinA/1sinA)

    =(1+sinA)²/1sin²A)+(1sin²A)/(1+sinA)²

    =1+sinA/cosA+cosA/1+sinA

    =(1+sinA)²+cos²A/cosA(1+sinA)

    =1+sin²A+cos²A+2sinA/cosA (1+sinA)

    =2+2sinA/cosA (1+sinA)

    =2(1+sinA)/cosA(1+sinA)

    =2/CosA

    =2SecA

    Hence proved.....

    Step-by-step explanation:

    Hope it helps you.......

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