Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If all the roots of the equation x3−3x=0x^{3}-3x=0 satisfy the equation (α−sin⁡−1(sin⁡2))x2−(β−tan⁡−1(tan⁡1))x+γ2−2γ+1=0(\alpha-\sin^{-1}(\sin 2))x^{2}-(\beta-\tan^{-1}(\tan 1))x+\gamma^{2}-2\gamma+1=0, then find the value of ∣cot⁡(β+γ)+cot⁡α∣|\cot(\beta+\gamma)+\cot\alpha|.
Solution
Answer: 0
Step 1: The roots of x3−3x=0x^{3}-3x=0 are x=0,3,−3x=0,\sqrt{3},-\sqrt{3} (three distinct values). Step 2: A quadratic with three distinct roots must be the zero polynomial. So every coefficient is zero:
α−sin⁡−1(sin⁡2)=0\alpha-\sin^{-1}(\sin 2)=0
β−tan⁡−1(tan⁡1)=0\beta-\tan^{-1}(\tan 1)=0
γ2−2γ+1=0\gamma^{2}-2\gamma+1=0
 Step 3: Compute each: For sin⁡−1(sin⁡2)\sin^{-1}(\sin 2): Since 2∈(π2,π)2\in\left(\dfrac{\pi}{2},\pi\right), sin⁡−1(sin⁡2)=π−2\sin^{-1}(\sin 2)=\pi-2. So α=π−2\alpha=\pi-2. For tan⁡−1(tan⁡1)\tan^{-1}(\tan 1): Since 1∈(−π2,π2)1\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right), tan⁡−1(tan⁡1)=1\tan^{-1}(\tan 1)=1. So β=1\beta=1. For γ\gamma: (γ−1)2=0⇒γ=1(\gamma-1)^{2}=0\Rightarrow\gamma=1. Step 4: Compute β+γ=1+1=2\beta+\gamma=1+1=2, so cot⁡(β+γ)=cot⁡2\cot(\beta+\gamma)=\cot 2. And cot⁡α=cot⁡(π−2)=−cot⁡2\cot\alpha=\cot(\pi-2)=-\cot 2. Step 5: Therefore:
∣cot⁡(β+γ)+cot⁡α∣=∣cot⁡2+(−cot⁡2)∣=0|\cot(\beta+\gamma)+\cot\alpha|=|\cot 2+(-\cot 2)|=0
 Answer: 00
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