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Exercise 4.6 · Q2

Q.If sin⁡−1x+sin⁡−1y=2π3\sin^{-1}x + \sin^{-1}y = \dfrac{2\pi}3; then cos⁡−1x+cos⁡−1y\cos^{-1}x + \cos^{-1}y is equal to

(1) 2π3\dfrac{2\pi}3
(2) π3\dfrac{\pi}3
(3) π6\dfrac{\pi}6
(4) π\pi
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✓ Free question

Rewriting each sin⁡−1\sin^{-1} term via the complementary identity turns the given equation directly into the target sum.

Step 1. Rewrite sin⁡−1x\sin^{-1}x and sin⁡−1y\sin^{-1}y. sin⁡−1x=π2−cos⁡−1x\sin^{-1}x=\dfrac{\pi}2-\cos^{-1}x and sin⁡−1y=π2−cos⁡−1y\sin^{-1}y=\dfrac{\pi}2-\cos^{-1}y.

Step 2. Substitute into the given equation. (π2−cos⁡−1x)+(π2−cos⁡−1y)=2π3⇒π−(cos⁡−1x+cos⁡−1y)=2π3\left(\dfrac{\pi}2-\cos^{-1}x\right)+\left(\dfrac{\pi}2-\cos^{-1}y\right)=\dfrac{2\pi}3\Rightarrow\pi-\left(\cos^{-1}x+\cos^{-1}y\right)=\dfrac{2\pi}3.

Step 3. Solve for the target sum. cos⁡−1x+cos⁡−1y=π−2π3=π3\cos^{-1}x+\cos^{-1}y=\pi-\dfrac{2\pi}3=\dfrac{\pi}3.

✓Final answer

Option (2): π3\dfrac{\pi}3.

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