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Question: If \(\sec \theta + \tan \theta = p\), obtain the value of \(\sec \theta ,\tan \theta ,\sin \theta \)...

If sec⁡θ+tan⁡θ=p\sec \theta + \tan \theta = p, obtain the value of sec⁡θ,tan⁡θ,sin⁡θ\sec \theta ,\tan \theta ,\sin \theta in terms of p.

Explanation

Solution

Hint: Here we use the trigonometry identities and formulae to obtain the values.

“Complete step-by-step answer:”
As we know sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1then
(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1 (sec⁡θ−tan⁡θ)=1(sec⁡θ+tan⁡θ) (sec⁡θ−tan⁡θ)=1p→(1) [∵(sec⁡θ−tan⁡θ)=p] (sec⁡θ+tan⁡θ)=p→(2)  (\sec \theta + \tan \theta )(\sec \theta - \tan \theta ) = 1 \\\ (\sec \theta - \tan \theta ) = \dfrac{1}{{(\sec \theta + \tan \theta )}} \\\ (\sec \theta - \tan \theta ) = \dfrac{1}{p} \to (1){\text{ [}}\because (\sec \theta - \tan \theta ) = p] \\\ (\sec \theta + \tan \theta ) = p \to (2) \\\
Now add equation (1) and (2) we get
sec⁡θ=12(p2+1p)\sec \theta = \dfrac{1}{2}\left( {\dfrac{{{p^2} + 1}}{p}} \right)
Now subtract equation (1) and (2) we get
tan⁡θ=12(p2−1p)\tan \theta = \dfrac{1}{2}\left( {\dfrac{{{p^2} - 1}}{p}} \right)
From the values of sec⁡θ\sec \theta andtan⁡θ\tan \theta , we know tan⁡θsec⁡θ=sin⁡θ\dfrac{{\tan \theta }}{{\sec \theta }} = \sin \theta
Therefore, sin⁡θ=p2−1p2+1\sin \theta = \dfrac{{{p^2} - 1}}{{{p^2} + 1}}.
Hence, we get the answer.

Note: Whenever such type of questions are always try to start the question with use of the trigonometric identities and use some algebraic formula to find the answer as we know (a2−b2)=(a+b)(a−b)({a^2} - {b^2}) = (a + b)(a - b) that we apply in the question to solve it.