if tan2⁡θ+sec⁡2θ=1, then θ=

A
$n π$
B
$n π \pm \frac{π}{6}$
C
$2 n π \pm \frac{π}{12}$
D
$2 n π \pm \frac{π}{4}$

Solution:

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

Let $t = \tan^{2} θ$ So, $(i)$ becomes $t + \frac{1 + t}{1 - t} - 1 = 0$

$\begin{array}{l} \Rightarrow t - t^{2} + 1 + t - 1 + t = 0 \Rightarrow t^{2} = 3 t \\ ∴ t = 0 \Rightarrow \tan^{2} θ = 0 \Rightarrow 0 = n π \\ \Rightarrow \tan θ = \pm \sqrt{3} \Rightarrow θ = n π \pm \frac{π}{3}, t = 3 \Rightarrow \tan^{2} θ = 3 \end{array}$

if $\tan^{2} θ + \sec 2 θ = 1,$ then $θ =$

Solution:

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