SecA tanB+tanA secB=2,secA secB+tanA tanB=k then k square=

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SecA tanB+tanA secB=2,secA secB+tanA tanB=k then k square=

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Jade 12 hours 2021-11-25T01:41:25+00:00 1 Answer 0 views 0

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    2021-11-25T01:42:41+00:00

    \textbf{Given:}

    secA\,tanB+tanA\,secB=2

    secA\,secB+tanA\,tanB=k

    \textbf{To find:}

    \text{The value of $k^2$}

    \textbf{Solution:}

    \text{Consider,}

    k^2

    =(secA\,secB+tanA\,tanB)^2

    =sec^2A\,sec^2B+tan^2A\,tan^2B+2\,secA\,secB\,tanA\,tanB

    =sec^2A(1+tan^2B)+tan^2A(sec^2B-1)+2\,secA\,secB\,tanA\,tanB

    =sec^2A+sec^2A\,tan^2B+tan^2A\,sec^2B-tan^2A+2\,secA\,secB\,tanA\,tanB

    =(sec^2A-tan^2A)+(sec^2A\,tan^2B+tan^2A\,sec^2B+2\,secA\,tanB\,tanA\,secB)

    =(sec^2A-tan^2A)+(secA\,tanB+tanA\,secB)^2

    =(1)+(2)^2

    =1+4

    =5

    \textbf{Answer:}

    \boxed{\textbf{The value of $\bf\,k^2$ is 5}}

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