if tan2⁡θ+sec⁡2θ=1, then θ=

# if ${\mathrm{tan}}^{2}\theta +\mathrm{sec}2\theta =1,$ then $\theta =$

1. A

$n\pi$

2. B

$n\pi ±\frac{\pi }{6}$

3. C

$2n\pi ±\frac{\pi }{12}$

4. D

$2n\pi ±\frac{\pi }{4}$

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### Solution:

${\mathrm{tan}}^{2}\theta +\frac{1}{\mathrm{cos}2\theta }-1=0$  …………(i)

Let $t={\mathrm{tan}}^{2}\theta$ So,  becomes $t+\frac{1+t}{1-t}-1=0$

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