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Question: If \(2 \tan ^ { 2 } \theta = \sec ^ { 2 } \theta\), then the general value of \(\theta\) is...

If 2tan⁡2θ=sec⁡2θ2 \tan ^ { 2 } \theta = \sec ^ { 2 } \theta, then the general value of θ\theta is

A

nπ+π4n \pi + \frac { \pi } { 4 }

B

nπ−π4n \pi - \frac { \pi } { 4 }

C

nπ±π4n \pi \pm \frac { \pi } { 4 }

D

2nπ±π42 n \pi \pm \frac { \pi } { 4 }

Answer

nπ±π4n \pi \pm \frac { \pi } { 4 }

Explanation

Solution

2tan⁡2θ=1+tan⁡2θ2 \tan ^ { 2 } \theta = 1 + \tan ^ { 2 } \theta ⇒ tan⁡2θ=1\tan ^ { 2 } \theta = 1 ⇒tan⁡2θ=tan⁡2π4\Rightarrow \tan ^ { 2 } \theta = \tan ^ { 2 } \frac { \pi } { 4 }

∴θ=nπ±π4\therefore \theta = n \pi \pm \frac { \pi } { 4 }