IfTantheta+ Sectheta √3, then theta=​

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IfTantheta+ Sectheta
√3, then theta=​

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Abigail 2 weeks 2021-09-14T01:53:08+00:00 1 Answer 0 views 0

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    2021-09-14T01:54:46+00:00

    Given,

    \longrightarrow\sec\theta+\tan\theta=\sqrt3\quad\quad\dots(1)

    We know that,

    \longrightarrow \sec^2\theta-\tan^2\theta=1

    \longrightarrow(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1

    From (1),

    \longrightarrow\sqrt3\,(\sec\theta-\tan\theta)=1

    \longrightarrow\sec\theta-\tan\theta=\dfrac{1}{\sqrt3}\quad\quad\dots(2)

    Adding (1) and (2),

    \longrightarrow(\sec\theta+\tan\theta)+(\sec\theta-\tan\theta)=\sqrt3+\dfrac{1}{\sqrt3}

    \longrightarrow2\sec\theta=\dfrac{4}{\sqrt3}

    \longrightarrow\sec\theta=\dfrac{2}{\sqrt3}

    \longrightarrow\sec\theta=\sec\left(\dfrac{\pi}{6}\right)

    Subtracting (2) from (1),

    \longrightarrow(\sec\theta+\tan\theta)-(\sec\theta-\tan\theta)=\sqrt3-\dfrac{1}{\sqrt3}

    \longrightarrow2\tan\theta=\dfrac{2}{\sqrt3}

    \longrightarrow\tan\theta=\dfrac{1}{\sqrt3}

    \longrightarrow\tan\theta=\tan\left(\dfrac{\pi}{6}\right)

    We get \theta lies in first quadrant since \sec\theta and \tan\theta are positive.

    So general value of \theta is,

    \longrightarrow\underline{\underline{\theta=2n\pi+\dfrac{\pi}{6}}},\quad n\in\mathbb{Z}

    And principal value of \theta is,

    \longrightarrow\underline{\underline{\theta=\dfrac{\pi}{6}}}

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