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If 4sin2θ=14sin^2\theta = 14sin2θ=1 and θ\thetaθ is an acute angle, then the value of cos2θ+tan2θcos^2\theta + tan^2\thetacos2θ+tan2θ&n
Question

If 4sin2θ=14sin^2\theta = 1 and θ\theta is an acute angle, then the value of cos2θ+tan2θcos^2\theta + tan^2\theta = _______.​

A.

1312\frac{13}{12}​​

B.

513\frac{5}{13}​​

C.

1

D.

14\frac{1}{4}​​

Correct option is A

Given: 

4sin2θ=14sin^2\theta = 1  

Solution:

4sin2θ=1sin2θ=14 sinθ=12 sinθ=sin30 θ=304sin^2\theta = 1 \\\sin^2 \theta = \frac{1}{4}\\\ \\sin\theta = \frac{1}{2}\\\ \\sin\theta = sin30^\circ \\\ \theta = 30^\circ 

Then the value of 

cos2θ+tan2θcos^2\theta + tan^2\theta    = cos²30° + tan²30° = 34+13=912+412=1312\frac{3}{4} + \frac{1}{3} = \frac{9}{12} + \frac{4}{12} = \frac{13}{12}​​

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