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Question: If \(\tan\frac{\theta}{2} = t,\) then \(\frac{1 - t^{2}}{1 + t^{2}}\) is equal to...

If tan⁡θ2=t,\tan\frac{\theta}{2} = t, then 1−t21+t2\frac{1 - t^{2}}{1 + t^{2}} is equal to

A

cos⁡θ\cos\theta

B

sin⁡θ\sin\theta

C

sec⁡θ\sec\theta

D

cos⁡2θ\cos 2\theta

Answer

cos⁡θ\cos\theta

Explanation

Solution

1−t21+t2\frac{1 - t^{2}}{1 + t^{2}}= 1−tan⁡2θ21+tan⁡2θ2\frac{1 - \tan^{2}\frac{\theta}{2}}{1 + \tan^{2}\frac{\theta}{2}} (∴ tan⁡θ/2=t)\tan\theta/2 = t) = cos⁡(2.θ/2)=cos⁡θ\cos(2.\theta/2) = \cos\theta.