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Miscellaneous Exercise 3 (I) · Q82

Q.Select the correct option: The value of cos⁡θ1+sin⁡θ\dfrac{\cos\theta}{1+\sin\theta} is equal to (A) tan⁡(θ2−π4)\tan\left(\dfrac{\theta}{2}-\dfrac{\pi}{4}\right) (B) −tan⁡(π4−θ2)-\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right) (C) tan⁡(π4−θ2)\tan\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right) (D) tan⁡(π4+θ2)\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)

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Step 1: cos⁡θ=(cos⁡θ2−sin⁡θ2)(cos⁡θ2+sin⁡θ2)\cos\theta=\left(\cos\dfrac{\theta}{2}-\sin\dfrac{\theta}{2}\right)\left(\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}\right) and 1+sin⁡θ=(sin⁡θ2+cos⁡θ2)21+\sin\theta=\left(\sin\dfrac{\theta}{2}+\cos\dfrac{\theta}{2}\right)^2.

Step 2: Cancel the common factor (cos⁡θ2+sin⁡θ2)\left(\cos\dfrac{\theta}{2}+\sin\dfrac{\theta}{2}\right): cos⁡θ1+sin⁡θ=cos⁡θ2−sin⁡θ2cos⁡θ2+sin⁡θ2\dfrac{\cos\theta}{1+\sin\theta}=\dfrac{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}. …

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