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Exercise 3.1 · Q9

Q.If sec⁡θ+tan⁡θ=p\sec\theta+\tan\theta=p, obtain the values of sec⁡θ\sec\theta, tan⁡θ\tan\theta and sin⁡θ\sin\theta in terms of pp.

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Factoring the Pythagorean identity sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1 as (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1 produces a second linear relation between sec⁡θ\sec\theta and tan⁡θ\tan\theta; combined with the given pp, this is a simple pair of simultaneous equations.

Step 1. Get a second relation from the Pythagorean identity. sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1 factors as (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1. Since sec⁡θ+tan⁡θ=p\sec\theta+\tan\theta=p,

sec⁡θ−tan⁡θ=1p.\sec\theta-\tan\theta=\frac{1}{p}.

Step 2. Solve for sec⁡θ\sec\theta. Adding sec⁡θ+tan⁡θ=p\sec\theta+\tan\theta=p and sec⁡θ−tan⁡θ=1p\sec\theta-\tan\theta=\dfrac1p:

2sec⁡θ=p+1p=p2+1p ⇒ sec⁡θ=p2+12p.2\sec\theta=p+\frac1p=\frac{p^2+1}{p} \ \Rightarrow\ \sec\theta=\frac{p^2+1}{2p}.

Step 3. Solve for tan⁡θ\tan\theta. Subtracting the two equations instead: …

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