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Inverse Trigonometric Functions question

2025 · 24 Jan · Shift 2 · Q32
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  5. /2025 · 24 Jan · Shift 2 · Q32

Inverse Trigonometric Functions question

2025 · 24 Jan · Shift 2 · Q32

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If α>β>γ>0\alpha\gt \beta\gt \gamma\gt 0α>β>γ>0, then the expression cot⁡−1{β+(1+β2)(α−β)}+cot⁡−1{γ+(1+γ2)(β−γ)}+cot⁡−1{α+(1+α2)(γ−α)}\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}cot−1{β+(α−β)(1+β2)​}+cot−1{γ+(β−γ)(1+γ2)​}+cot−1{α+(γ−α)(1+α2)​} is equal to :
  1. A
    3π3 \pi3π
  2. B
    π2−(α+β+γ)\frac{\pi}{2}-(\alpha+\beta+\gamma)2π​−(α+β+γ)
  3. C
    π\piπ
  4. D
    0
View written solutionFree

Correct answer: C

Let S=cot⁡−1(β+1+β2α−β)+cot⁡−1(γ+1+γ2β−γ)+cot⁡−1(α+1+α2γ−α).S=\cot^{-1}\left(\beta+\frac{1+\beta^2}{\alpha-\beta}\right)+\cot^{-1}\left(\gamma+\frac{1+\gamma^2}{\beta-\gamma}\right)+\cot^{-1}\left(\alpha+\frac{1+\alpha^2}{\gamma-\alpha}\right).S=cot−1(β+α−β1+β2​)+cot−1(γ+β−γ1+γ2​)+cot−1(α+γ−α1+α2​).

We simplify each term.

1. First term

Consider β+1+β2α−β.\beta+\frac{1+\beta^2}{\alpha-\beta}.β+α−β1+β2​. Taking LCM,

\frac{\beta(\alpha-\beta)+(1+\beta^2)}{\alpha-\beta} =\frac{\alpha\beta-\beta^2+1+\beta^2}{\alpha-\beta} =\frac{1+\alpha\beta}{\alpha-\beta}.$$ Now use the identity $$\cot(A-B)=\frac{1+\cot A\cot B}{\cot B-\cot A}.$$ If we put $\cot A=\alpha$ and $\cot B=\beta$, then $$\cot(B-A)=\frac{1+\alpha\beta}{\alpha-\beta}.$$ Hence $$\cot^{-1}\left(\beta+\frac{1+\beta^2}{\alpha-\beta}\right)=\cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right).$$ Since $\alpha>\beta>0$, we have $\alpha-\beta>0$, so the argument is positive. Therefore this principal value lies in $(0,\pi/2)$, and we may write $$\cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right)=B-A,$$ where $A=\cot^{-1}\alpha$, $B=\cot^{-1}\beta$. Thus $$T_1=\cot^{-1}\beta-\cot^{-1}\alpha.$$ ## 2. Second term Similarly, $$\gamma+\frac{1+\gamma^2}{\beta-\gamma} =\frac{\gamma(\beta-\gamma)+1+\gamma^2}{\beta-\gamma} =\frac{1+\beta\gamma}{\beta-\gamma}.$$ Using the same identity with $\cot A=\beta$, $\cot B=\gamma$, $$\frac{1+\beta\gamma}{\beta-\gamma}=\cot(\cot^{-1}\gamma-\cot^{-1}\beta).$$ Since $\beta>\gamma>0$, this is positive, so $$T_2=\cot^{-1}\gamma-\cot^{-1}\beta.$$ ## 3. Third term Now $$\alpha+\frac{1+\alpha^2}{\gamma-\alpha} =\frac{\alpha(\gamma-\alpha)+1+\alpha^2}{\gamma-\alpha} =\frac{1+\alpha\gamma}{\gamma-\alpha}.

Since α>γ\alpha>\gammaα>γ, we have γ−α<0\gamma-\alpha<0γ−α<0, so this quantity is negative.

Also, 1+αγγ−α=cot⁡(cot⁡−1α−cot⁡−1γ).\frac{1+\alpha\gamma}{\gamma-\alpha}=\cot(\cot^{-1}\alpha-\cot^{-1}\gamma).γ−α1+αγ​=cot(cot−1α−cot−1γ). Let A=cot⁡−1α,G=cot⁡−1γ.A=\cot^{-1}\alpha,\quad G=\cot^{-1}\gamma.A=cot−1α,G=cot−1γ. Because α>γ>0\alpha>\gamma>0α>γ>0, we get 0<A<G<π2,0<A<G<\frac{\pi}{2},0<A<G<2π​, so A−G<0.A-G<0.A−G<0. To convert to principal value of cot⁡−1\cot^{-1}cot−1, note that for a negative cotangent value, the angle lies in (π/2,π)(\pi/2,\pi)(π/2,π). Hence T3=π+(A−G)=π+cot⁡−1α−cot⁡−1γ.T_3=\pi+(A-G)=\pi+\cot^{-1}\alpha-\cot^{-1}\gamma.T3​=π+(A−G)=π+cot−1α−cot−1γ.

4. Add all three terms

Therefore, S=(cot⁡−1β−cot⁡−1α)+(cot⁡−1γ−cot⁡−1β)+(π+cot⁡−1α−cot⁡−1γ).S=(\cot^{-1}\beta-\cot^{-1}\alpha)+(\cot^{-1}\gamma-\cot^{-1}\beta)+\left(\pi+\cot^{-1}\alpha-\cot^{-1}\gamma\right).S=(cot−1β−cot−1α)+(cot−1γ−cot−1β)+(π+cot−1α−cot−1γ).

All variable terms cancel: S=π.S=\pi.S=π.

5. Check options

Thus the expression is π.\boxed{\pi}.π​. So the correct option is:

C: π\piπ

6. Comparison with stored answer

Stored correct answer: C

Our derived answer also gives C. Hence they agree.

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