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Exercises · Q8

Q.If sec⁡θ+tan⁡θ=5\sec\theta+\tan\theta=5, find the value of sec⁡θ−tan⁡θ\sec\theta-\tan\theta.

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Step 1 — recall the identity. 1+tan⁡2θ=sec⁡2θ⇒sec⁡2θ−tan⁡2θ=11+\tan^2\theta=\sec^2\theta \Rightarrow \sec^2\theta-\tan^2\theta=1.

Step 2 — factorise the left side as a difference of squares. sec⁡2θ−tan⁡2θ=(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)\sec^2\theta-\tan^2\theta=(\sec\theta+\tan\theta)(\sec\theta-\tan\theta), so (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1.

Step 3 — substitute the given value. 5×(sec⁡θ−tan⁡θ)=1⇒sec⁡θ−tan⁡θ=155\times(\sec\theta-\tan\theta)=1 \Rightarrow \sec\theta-\tan\theta=\dfrac15. …

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