If sec⁡θ+tan⁡θ=x, obtain the values of sec⁡θ,tan⁡θ and sin⁡θ.

We have to use properties of trigonometry. (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)(sec⁡θ−tan⁡θ)=x →sec2θ−tan2θsec⁡θ−tan⁡θ=x →sec⁡θ−tan⁡θ=1x and sec⁡θ+tan⁡θ=x sec⁡θ=12(x+1x) tan⁡θ=12(x−1x) →1cos⁡θ=12(1+x2x) →cos⁡θ=2x1+x2 sin⁡θ=1−4x2(1+x2)2=(x2−1)2(x2+1)2 sin⁡θ=x2−1x2+1

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rocedwrp

Answered question

2021-08-13

\sec θ + \tan θ = x

, obtain the values of

\sec θ, \tan θ

and

\sin θ

Answer & Explanation

un4t5o4v

Skilled2021-08-14Added 105 answers

We have to use properties of trigonometry.

(\sec θ + \tan θ) \frac{(\sec θ - \tan θ)}{(\sec θ - \tan θ)} = x

\to \frac{\sec^{2} θ - \tan^{2} θ}{\sec θ - \tan θ} = x

\to \sec θ - \tan θ = \frac{1}{x}

and

\sec θ + \tan θ = x

\sec θ = \frac{1}{2} (x + \frac{1}{x})

\tan θ = \frac{1}{2} (x - \frac{1}{x})

\to \frac{1}{\cos θ} = \frac{1}{2} (\frac{1 + x^{2}}{x})

\to \cos θ = \frac{2 x}{1 + x^{2}}

\sin θ = \sqrt{1 - \frac{4 x^{2}}{{(1 + x^{2})}^{2}}} = \sqrt{\frac{{(x^{2} - 1)}^{2}}{{(x^{2} + 1)}^{2}}}

\sin θ = \frac{x^{2} - 1}{x^{2} + 1}

Jeffrey Jordon

Expert2021-12-10Added 2605 answers

Answer is given below (on video)

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