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Mathematics Question on Trigonometric Functions

The value of cos⁡θ1+sin⁡θ\frac{\cos\theta}{1+ \sin\theta} is equal to

A

tan⁡(θ2−π4)\tan\left(\frac{\theta}{2} - \frac{\pi}{4}\right)

B

tan⁡(−π4−θ2)\tan\left( - \frac{\pi}{4}- \frac{\theta}{2} \right)

C

tan⁡(π4−θ2)\tan\left( \frac{\pi}{4}- \frac{\theta}{2} \right)

D

tan⁡(π4+θ2)\tan\left( \frac{\pi}{4} + \frac{\theta}{2} \right)

Answer

tan⁡(π4−θ2)\tan\left( \frac{\pi}{4}- \frac{\theta}{2} \right)

Explanation

Solution

cos⁡θ1+sin⁡θ=sin⁡(π2−θ)1+cos⁡(π2−θ)\frac{\cos\theta}{1+ \sin\theta} = \frac{ \sin\left(\frac{\pi}{2} - \theta\right)}{1 + \cos \left(\frac{\pi}{2} - \theta\right)}
=2sin⁡(π4−θ2)cos⁡(π4−θ2)2cos⁡2(π4−θ2)= \frac{2 \sin\left( \frac{\pi}{4} - \frac{\theta}{2}\right)\cos \left(\frac{\pi}{4} - \frac{\theta}{2}\right)}{2 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)}
=sin⁡(π4−θ2)cos⁡(π4−θ2)= \frac{\sin\left(\frac{\pi}{4} - \frac{\theta}{2}\right)}{\cos\left(\frac{\pi}{4} - \frac{\theta}{2}\right)}
=tan⁡(π4−θ2)= \tan \left(\frac{\pi}{4} - \frac{\theta}{2}\right)
cos⁡θ1+sin⁡θ\frac{\cos \theta}{1+\sin \theta}
=(cos⁡2θ/2−sin⁡2θ/2)(sin⁡2θ/2+cos⁡2θ/2)+2sin⁡θ/2+cos⁡θ/2=\frac{\left(\cos ^{2} \theta / 2-\sin ^{2} \theta / 2\right)}{\left(\sin ^{2} \theta / 2+\cos ^{2} \theta / 2\right)+2 \sin \theta / 2+\cos \theta / 2}
=(cos⁡θ/2+sin⁡θ/2)(cos⁡θ/2−sin⁡θ/2)(cos⁡θ/2+sin⁡θ/2)2=\frac{(\cos \theta / 2+\sin \theta / 2)(\cos \theta / 2-\sin \theta / 2)}{(\cos \theta / 2+\sin \theta / 2)^{2}}
=1−tan⁡θ/21+tan⁡θ/2=\frac{1-\tan \theta / 2}{1+\tan \theta / 2}
=tan⁡π/4−tan⁡θ/21+tan⁡π/4⋅tan⁡θ/2=\frac{\tan \pi / 4-\tan \theta / 2}{1+\tan \pi / 4 \cdot \tan \theta / 2}
=tan⁡(π4−θ2)=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)