If sec⁡θtan⁡θ=2, then θ= - Infinity Learn by Sri Chaitanya

A
$n π + (- 1)^{n} \frac{π}{4}, n \equiv Z$
B
$2 n π \pm \frac{π}{3}, n \in Z$
C
$n π \pm \frac{2 π}{3}, n \in Z$
D
$n π - \frac{π}{4}, n \in Z$

Solution:

We have

$\begin{array}{l} = \sec θ \tan θ = \sqrt{2} \\ = \sin θ = \sqrt{2} \cos^{2} θ \\ = \sin θ = \sqrt{2} - \sqrt{2} \sin^{2} θ \\ = \sqrt{2} \sin^{2} θ + \sin θ - \sqrt{2} = 0 \end{array}$

$\begin{array}{l} \Rightarrow (\sqrt{2} \sin θ - 1) (\sin θ + \sqrt{2}) = 0 \\ \Rightarrow \sqrt{2} \sin θ - 1 = 0 \\ \Rightarrow \sin θ = \frac{1}{\sqrt{2}} = \sin \frac{π}{4} \Rightarrow θ = n π + (- 1)^{n} \frac{π}{4}, n \in Z \end{array}$

If $\sec θ \tan θ = \sqrt{2}, then θ =$

Solution:

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