Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

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

Question
If the roots of the equation x3−10x+11=0x^{3}-10x+11=0 are u,v,wu,v,w, then find the value of 3csc⁡2(tan⁡−1u+tan⁡−1v+tan⁡−1w)3\csc^{2}(\tan^{-1}u+\tan^{-1}v+\tan^{-1}w).
Solution
Answer: 6
Step 1: By Vieta's formulas for x3+0⋅x2−10x+11=0x^{3}+0\cdot x^{2}-10x+11=0:
S1=u+v+w=0S_{1}=u+v+w=0
S2=uv+vw+wu=−10S_{2}=uv+vw+wu=-10
S3=uvw=−11S_{3}=uvw=-11
 Step 2: Let α=tan⁡−1u\alpha=\tan^{-1}u, β=tan⁡−1v\beta=\tan^{-1}v, γ=tan⁡−1w\gamma=\tan^{-1}w, so tan⁡α=u\tan\alpha=u, tan⁡β=v\tan\beta=v, tan⁡γ=w\tan\gamma=w. Step 3: Use the identity:
tan⁡(α+β+γ)=S1−S31−S2=0−(−11)1−(−10)=1111=1\tan(\alpha+\beta+\gamma)=\dfrac{S_{1}-S_{3}}{1-S_{2}}=\dfrac{0-(-11)}{1-(-10)}=\dfrac{11}{11}=1
 Step 4: Therefore:
α+β+γ=tan⁡−11=π4\alpha+\beta+\gamma=\tan^{-1}1=\dfrac{\pi}{4}
 Step 5: Hence:
3csc⁡2(α+β+γ)=3csc⁡2π4=3⋅(2)2=3⋅2=63\csc^{2}(\alpha+\beta+\gamma)=3\csc^{2}\dfrac{\pi}{4}=3\cdot(\sqrt{2})^{2}=3\cdot 2=6
 Answer: 66
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