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    ​If (1+tana)(1+tan 4a)=2,a ∈(0,π16)(1+\text{tan}a)(1+\text{tan}\space4a)=2, a\space∈(0,\frac{π}{16})(1+tana)(1+tan 4a)=2,a ∈(
    Question

    If (1+tana)(1+tan 4a)=2,a (0,π16)(1+\text{tan}a)(1+\text{tan}\space4a)=2, a\space∈(0,\frac{π}{16})​ then aa is equal to:

    A.

    π40\frac{π}{40}​​

    B.

    π60\frac{π}{60}​​

    C.

    π20\frac{π}{20}​​

    D.

    π30\frac{π}{30}​​

    Correct option is C

    Given:
    The equation:
    (1+tana)(1+tan4a)=2(1 + \tan a)(1 + \tan 4a) = 2​​
    where a (0,π16)\in \left(0, \frac{\pi}{16}\right)
    Concept Used:

    (1 + tanA)(1 + tanB) =2

    then, A + B =π4\frac{\pi}{4}​​

    Solution:

    (1+tana)(1+tan4a)=2(1 + \tan a)(1 + \tan 4a) = 2

    aa​+ 4aa​ = π4\frac{\pi}{4}

    5 aa​=π4\frac{\pi}{4}

    aa​ =π20\frac{\pi}{20}(0,π16)\in \left(0, \frac{\pi}{16}\right) ​​​


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