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Question: If \(\tan\theta + \sin\theta = m\)and \(\tan\theta - \sin\theta = n,\) then...

If tan⁡θ+sin⁡θ=m\tan\theta + \sin\theta = mand tan⁡θ−sin⁡θ=n,\tan\theta - \sin\theta = n, then

A

m2−n2=4mnm^{2} - n^{2} = 4mn

B

m2+n2=4mnm^{2} + n^{2} = 4mn

C

m2−n2=m2+n2m^{2} - n^{2} = m^{2} + n^{2}

D

m2−n2=4mnm^{2} - n^{2} = 4\sqrt{mn}

Answer

m2−n2=4mnm^{2} - n^{2} = 4\sqrt{mn}

Explanation

Solution

(m+n)=2tan⁡θ,m−n=2sin⁡θ(m + n) = 2\tan\theta,m - n = 2\sin\theta

∴m2−n2=4tan⁡θ.sin⁡θ\therefore m^{2} - n^{2} = 4\tan\theta.\sin\theta …..(i)

4mn=4tan⁡2θ−sin⁡2θ=4sin⁡θ.tan⁡θ4\sqrt{mn} = 4\sqrt{\tan^{2}\theta - \sin^{2}\theta} = 4\sin\theta.\tan\theta …..(ii)

From (i) and (ii), m2−n2=4mnm^{2} - n^{2} = 4\sqrt{mn}.