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Question

Mathematics Question on Trigonometric Functions

If tan⁡ θ+sec θ=p\tan \, \theta + sec \, \theta = p, then what is the value of sec⁡ θ\sec \, \theta ?

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

Given that tan⁡ θ+sec⁡ θ=p\tan \, \theta + \sec \, \theta = p ....(1) and we know that ⇒ sec⁡2θ−tan⁡2θ=(sec⁡θ−tan⁡θ)p\Rightarrow \, \sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta) p (multiplying both the sides by (sec⁡θ−tan⁡θ)(\sec \theta - \tan \theta)) ⇒ (sec⁡θ−tan⁡θ)p=1\Rightarrow \, (\sec \theta - \tan \theta ) p =1 ⇒ sec⁡ θ−tan⁡ θ=1p\Rightarrow \, \sec \, \theta - \tan \, \theta = \frac{1}{p} .....(2) On solving equation (1) and (2), we get 2sec⁡θ=p2+1p ⇒ sec⁡θ=p2+12p2 \sec \theta = \frac{p^2 + 1}{p} \, \Rightarrow \, \sec \theta = \frac{p^2 + 1}{2p}