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Exercise 4.6 · Q20

Q.sin⁡(tan⁡−1x)\sin(\tan^{-1}x), ∣x∣<1|x| < 1 is equal to

(1) x1−x2\dfrac{x}{\sqrt{1-x^2}}
(2) 11−x2\dfrac1{\sqrt{1-x^2}}
(3) 11+x2\dfrac1{\sqrt{1+x^2}}
(4) x1+x2\dfrac{x}{\sqrt{1+x^2}}
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Naming θ=tan⁡−1x\theta=\tan^{-1}x and building its reference right triangle directly gives sin⁡θ\sin\theta as a ratio of xx and 1+x2\sqrt{1+x^2}.

Step 1. Let θ=tan⁡−1x\theta=\tan^{-1}x, so tan⁡θ=x\tan\theta=x, with θ∈(−π2,π2)\theta\in\left(-\dfrac{\pi}2,\dfrac{\pi}2\right).

Step 2. Build the reference triangle. Opposite side =x=x, adjacent side =1=1, hypotenuse =1+x2=\sqrt{1+x^2} (Pythagoras).

Step 3. Read off sin⁡θ\sin\theta. sin⁡θ=oppositehypotenuse=x1+x2\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{x}{\sqrt{1+x^2}}. …

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