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  4. If sin2 θ + 3 cos θ - 2 = 0, then cos3 θ + sec3 θ is equal to

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Q. If $\sin^2 \, \theta + 3 \, \cos \, \theta - 2 = 0$, then $\cos^3 \, \theta + \sec^3 \, \theta$ is equal to

Trigonometric Functions

Solution:

Given $\sin^2 \, \theta + 3 \, \cos \, \theta - 2 = 0$
$ \Rightarrow \cos^{2} \theta - 3 \cos \theta + 1 = 0 $
$\Rightarrow \cos^{2} \theta +1 = 3 \cos\theta $
$\Rightarrow \cos\theta + \frac{1}{\cos\theta} = 3 ,$
cubing both sides, we get
$ \cos^{3} \theta + \frac{1}{\cos^{3} \theta} + 3 \left(\cos \theta + \frac{1}{\cos\theta} \right) = 27$
$ \Rightarrow \cos^{3} \theta + \sec^{3} \theta = 27 - 3 \times3 = 18 $