if tan2⁡θ+sec⁡2θ=1, then θ=

if tan2θ+sec2θ=1, then θ=

  1. A

    nπ

  2. B

    nπ±π6

  3. C

    2nπ±π12

  4. D

    2nπ±π4

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    Solution:

    tan2θ+1cos2θ1=0  …………(i)

    Let t=tan2θ So,  (i)  becomes t+1+t1t1=0

    tt2+1+t1+t=0t2=3t t=0tan2θ=00=nπtanθ=±3θ=nπ±π3,t=3tan2θ=3

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