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Mathematics Question on Trigonometric Functions

If θ∈(π2,3π2)\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), then the value of 4cos⁡4θ+sin⁡22θ+4cot⁡θcos⁡2(π4−θ2)\sqrt{4 \cos ^{4} \theta+\sin ^{2} 2 \theta}+4 \cot \theta \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)

A

−2cot⁡θ-2 \cot \theta

B

2cot⁡θ2 \cot \theta

C

2cos⁡θ2 \cos \theta

D

2sin⁡θ2 \sin \theta

Answer

2cot⁡θ2 \cot \theta

Explanation

Solution

4cos⁡4θ+sin⁡22θ+4cot⁡θcos⁡2(π4−θ2)\sqrt{4 \cos ^{4} \theta+\sin ^{2} 2 \theta}+4 \cot \theta \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)
=4cos⁡4θ+(2sin⁡θcos⁡θ)2=\sqrt{4 \cos ^{4} \theta+(2 \sin \theta \cos \theta)^{2}}
+2cot⁡θ[2cos⁡2(π4−θ2)]+2 \cot \theta\left[2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]
=4cos⁡4θ+4sin⁡2θcos⁡2θ=\sqrt{4 \cos ^{4} \theta+4 \sin ^{2} \theta \cos ^{2} \theta}
+2cot⁡θ[1+cos⁡(π2−θ)]+2 \cot \theta\left[1+\cos \left(\frac{\pi}{2}-\theta\right)\right]
[∵2cos⁡2θ=1+cos⁡2θ]\left[\because 2 \cos ^{2} \theta=1+\cos 2 \theta\right]
=4cos⁡2θ(cos⁡2θ+sin⁡2θ)+2cot⁡θ(1+sin⁡θ)=\sqrt{4 \cos ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}+2 \cot \theta(1+\sin \theta)
=∣2cos⁡θ∣+2cot⁡θ+2cos⁡θ=|2 \cos \theta|+2 \cot \theta+2 \cos \theta
=−2cos⁡θ+2cot⁡θ+2cos⁡θ=-2 \cos \theta+2 \cot \theta+2 \cos \theta
[forθ∈(π2,3π2),cos⁡θ∣=−cos⁡θ]\left[\text{for} \theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), \cos \theta \mid=-\cos \theta\right]
=2cot⁡θ=2 \cot \theta