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Question 139 of 175

Q.If tan⁡θ+sin⁡θ=p\tan\theta + \sin\theta = p, tan⁡θ−sin⁡θ=q\tan\theta - \sin\theta = q and p>qp > q then show that p2−q2=4pqp^2 - q^2 = 4\sqrt{pq}.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2018Subjective· 3mImportance★★★★★
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Both p2−q2p^2-q^2 and 4pq4\sqrt{pq} simplify to 4sin⁡2θcos⁡θ\dfrac{4\sin^2\theta}{\cos\theta}, proving the identity.

Given p=tan⁡θ+sin⁡θp=\tan\theta+\sin\theta and q=tan⁡θ−sin⁡θq=\tan\theta-\sin\theta.

Compute p2−q2p^2-q^2: using (p+q)(p−q)(p+q)(p-q):

p+q=2tan⁡θp+q = 2\tan\theta, p−q=2sin⁡θp-q = 2\sin\theta

p2−q2=(2tan⁡θ)(2sin⁡θ)=4tan⁡θsin⁡θ=4⋅sin⁡θcos⁡θ⋅sin⁡θ=4sin⁡2θcos⁡θp^2-q^2 = (2\tan\theta)(2\sin\theta) = 4\tan\theta\sin\theta = 4\cdot\dfrac{\sin\theta}{\cos\theta}\cdot\sin\theta = \dfrac{4\sin^2\theta}{\cos\theta}

Compute pqpq:

pq=(tan⁡θ+sin⁡θ)(tan⁡θ−sin⁡θ)=tan⁡2θ−sin⁡2θpq = (\tan\theta+\sin\theta)(\tan\theta-\sin\theta) = \tan^2\theta-\sin^2\theta

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