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Question: If \(\sec\theta + \tan\theta = p,\) then \(\tan\theta\)is equal to...

If sec⁡θ+tan⁡θ=p,\sec\theta + \tan\theta = p, then tan⁡θ\tan\thetais equal to

A

2pp2−1\frac{2p}{p^{2} - 1}

B

p2−12p\frac{p^{2} - 1}{2p}

C

p2+12p\frac{p^{2} + 1}{2p}

D

2pp2+1\frac{2p}{p^{2} + 1}

Answer

p2−12p\frac{p^{2} - 1}{2p}

Explanation

Solution

sec⁡θ+tan⁡θ=p⇒sec⁡θ−tan⁡θ=1p\sec\theta + \tan\theta = p \Rightarrow \sec\theta - \tan\theta = \frac{1}{p}

Subtracting second from first, we get 2tan⁡θ=p−1p2\tan\theta = p - \frac{1}{p}

⇒tan⁡θ=p2−12p\Rightarrow \tan\theta = \frac{p^{2} - 1}{2p}.