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Exercise 4.6 · Q12

Q.If cot⁡−12\cot^{-1}2 and cot⁡−13\cot^{-1}3 are two angles of a triangle, then the third angle is

(1) π4\dfrac{\pi}4
(2) 3π4\dfrac{3\pi}4
(3) π6\dfrac{\pi}6
(4) π3\dfrac{\pi}3
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The three angles of a triangle sum to π\pi; converting the two given cot⁡−1\cot^{-1} angles to tan⁡−1\tan^{-1} form lets us add them with the standard sum formula before subtracting from π\pi.

Step 1. Convert to tan⁡−1\tan^{-1} form. cot⁡−12=tan⁡−112\cot^{-1}2=\tan^{-1}\dfrac12 and cot⁡−13=tan⁡−113\cot^{-1}3=\tan^{-1}\dfrac13 (both positive, so the reciprocal conversion is direct).

Step 2. Add via the tangent sum formula. 12+13=56\dfrac12+\dfrac13=\dfrac56; 12⋅13=16\dfrac12\cdot\dfrac13=\dfrac16, so 1−16=561-\dfrac16=\dfrac56. Ratio =5/65/6=1=\dfrac{5/6}{5/6}=1. …

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