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Exercise 4.6 · Q18

Q.If sin⁡−1x+cot⁡−1(12)=π2\sin^{-1}x + \cot^{-1}\left(\dfrac12\right) = \dfrac{\pi}2, then xx is equal to

(1) 12\dfrac12
(2) 15\dfrac1{\sqrt5}
(3) 25\dfrac2{\sqrt5}
(4) 32\dfrac{\sqrt3}2
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Rewriting cot⁡−1(12)\cot^{-1}\left(\dfrac12\right) as π2−tan⁡−1(12)\dfrac{\pi}2-\tan^{-1}\left(\dfrac12\right) isolates sin⁡−1x=tan⁡−1(12)\sin^{-1}x=\tan^{-1}\left(\dfrac12\right), and a 11–22–5\sqrt5 reference triangle gives xx.

Step 1. Use the complementary identity for t=12t=\dfrac12. tan⁡−112+cot⁡−112=π2⇒cot⁡−112=π2−tan⁡−112\tan^{-1}\dfrac12+\cot^{-1}\dfrac12=\dfrac{\pi}2\Rightarrow\cot^{-1}\dfrac12=\dfrac{\pi}2-\tan^{-1}\dfrac12.

Step 2. Substitute into the given equation. sin⁡−1x+(π2−tan⁡−112)=π2⇒sin⁡−1x=tan⁡−112\sin^{-1}x+\left(\dfrac{\pi}2-\tan^{-1}\dfrac12\right)=\dfrac{\pi}2\Rightarrow\sin^{-1}x=\tan^{-1}\dfrac12. …

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