## Problem Identification **Subject:** Mathematics (Trigonometry) **Core topic:** Simplifying inverse cotangent expressions using identities like \[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \] and principal values of $\cot^{-1}x$. --- ## Analysis ### Given \[ \alpha>\beta>\gamma>0 \] We need to evaluate: \[ S=\cot^{-1}\left\{\beta+\frac{1+\beta^2}{\alpha-\beta}\right\}+\cot^{-1}\left\{\gamma+\frac{1+\gamma^2}{\beta-\gamma}\right\}+\cot^{-1}\left\{\alpha+\frac{1+\alpha^2}{\gamma-\alpha}\right\} \] ### What to find Compute the exact value of $S$. ### Key ideas used 1. Algebraic simplification inside each $\cot^{-1}$. 2. For $x>0$: \[ \cot^{-1}x=\tan^{-1}\left(\frac{1}{x}\right) \] 3. For $K>0$, since $\cot(\pi-\theta)=-\cot\theta$ and principal range of $\cot^{-1}$ is $(0,\pi)$: \[ \cot^{-1}(-K)=\pi-\cot^{-1}(K) \] 4. If $A=\tan^{-1}\alpha$, $B=\tan^{-1}\beta$, then (with correct range here): \[ \tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)=A-B \] --- ## Solution Steps ### 1) Simplify the first term Inside: \[ \beta+\frac{1+\beta^2}{\alpha-\beta} =\frac{\beta(\alpha-\beta)+(1+\beta^2)}{\alpha-\beta} =\frac{\alpha\beta-\beta^2+1+\beta^2}{\alpha-\beta} =\frac{1+\alpha\beta}{\alpha-\beta} \] So, \[ \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) \] Since $\alpha>\beta>0$, we have $\alpha-\beta>0$ and $1+\alpha\beta>0$, so the argument is positive. Hence: \[ \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) =\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right) \] --- ### 2) Simplify the second term Similarly: \[ \gamma+\frac{1+\gamma^2}{\beta-\gamma} =\frac{\gamma(\beta-\gamma)+1+\gamma^2}{\beta-\gamma} =\frac{\beta\gamma-\gamma^2+1+\gamma^2}{\beta-\gamma} =\frac{1+\beta\gamma}{\beta-\gamma} \] Thus: \[ \cot^{-1}\left(\frac{1+\beta\gamma}{\beta-\gamma}\right) =\tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right) \] because $\beta-\gamma>0$. --- ### 3) Simplify the third term (sign matters) Inside: \[ \alpha+\frac{1+\alpha^2}{\gamma-\alpha} =\frac{\alpha(\gamma-\alpha)+1+\alpha^2}{\gamma-\alpha} =\frac{\alpha\gamma-\alpha^2+1+\alpha^2}{\gamma-\alpha} =\frac{1+\alpha\gamma}{\gamma-\alpha} \] But $\gamma-\alpha<0$ and $1+\alpha\gamma>0$, so the whole fraction is **negative**. Write: \[ \frac{1+\alpha\gamma}{\gamma-\alpha}=-\frac{1+\alpha\gamma}{\alpha-\gamma}=-K \] where \[ K=\frac{1+\alpha\gamma}{\alpha-\gamma}>0 \] So the third term is: \[ \cot^{-1}(-K)=\pi-\cot^{-1}(K) \] and since $K>0$, \[ \cot^{-1}(K)=\tan^{-1}\left(\frac{1}{K}\right)=\tan^{-1}\left(\frac{\alpha-\gamma}{1+\alpha\gamma}\right) \] Hence third term becomes: \[ \pi-\tan^{-1}\left(\frac{\alpha-\gamma}{1+\alpha\gamma}\right) =\pi+\tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right) \] --- ### 4) Add all three terms So \[ S=\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)+\tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right)+\pi+\tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right) \] Now set: \[ A=\tan^{-1}\alpha,\quad B=\tan^{-1}\beta,\quad C=\tan^{-1}\gamma \] Since $\alpha>\beta>\gamma>0$, we get: \[ 0\beta>\gamma$ but $\gamma<0$, which step in the solution breaks first and why (hint: principal values/sign of the third term)?## Problem Identification **Subject:** Mathematics (Trigonometry) **Core topic:** Simplifying inverse cotangent expressions using identities like \[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \] and principal values of $\cot^{-1}x$. --- ## Analysis ### Given \[ \alpha>\beta>\gamma>0 \] We need to evaluate: \[ S=\cot^{-1}\left\{\beta+\frac{1+\beta^2}{\alpha-\beta}\right\}+\cot^{-1}\left\{\gamma+\frac{1+\gamma^2}{\beta-\gamma}\right\}+\cot^{-1}\left\{\alpha+\frac{1+\alpha^2}{\gamma-\alpha}\right\} \] ### What to find Compute the exact value of $S$. ### Key ideas used 1. Algebraic simplification inside each $\cot^{-1}$. 2. For $x>0$: \[ \cot^{-1}x=\tan^{-1}\left(\frac{1}{x}\right) \] 3. For $K>0$, since $\cot(\pi-\theta)=-\cot\theta$ and principal range of $\cot^{-1}$ is $(0,\pi)$: \[ \cot^{-1}(-K)=\pi-\cot^{-1}(K) \] 4. If $A=\tan^{-1}\alpha$, $B=\tan^{-1}\beta$, then (with correct range here): \[ \tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)=A-B \] --- ## Solution Steps ### 1) Simplify the first term Inside: \[ \beta+\frac{1+\beta^2}{\alpha-\beta} =\frac{\beta(\alpha-\beta)+(1+\beta^2)}{\alpha-\beta} =\frac{\alpha\beta-\beta^2+1+\beta^2}{\alpha-\beta} =\frac{1+\alpha\beta}{\alpha-\beta} \] So, \[ \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) \] Since $\alpha>\beta>0$, we have $\alpha-\beta>0$ and $1+\alpha\beta>0$, so the argument is positive. Hence: \[ \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) =\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right) \] --- ### 2) Simplify the second term Similarly: \[ \gamma+\frac{1+\gamma^2}{\beta-\gamma} =\frac{\gamma(\beta-\gamma)+1+\gamma^2}{\beta-\gamma} =\frac{\beta\gamma-\gamma^2+1+\gamma^2}{\beta-\gamma} =\frac{1+\beta\gamma}{\beta-\gamma} \] Thus: \[ \cot^{-1}\left(\frac{1+\beta\gamma}{\beta-\gamma}\right) =\tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right) \] because $\beta-\gamma>0$. --- ### 3) Simplify the third term (sign matters) Inside: \[ \alpha+\frac{1+\alpha^2}{\gamma-\alpha} =\frac{\alpha(\gamma-\alpha)+1+\alpha^2}{\gamma-\alpha} =\frac{\alpha\gamma-\alpha^2+1+\alpha^2}{\gamma-\alpha} =\frac{1+\alpha\gamma}{\gamma-\alpha} \] But $\gamma-\alpha<0$ and $1+\alpha\gamma>0$, so the whole fraction is **negative**. Write: \[ \frac{1+\alpha\gamma}{\gamma-\alpha}=-\frac{1+\alpha\gamma}{\alpha-\gamma}=-K \] where \[ K=\frac{1+\alpha\gamma}{\alpha-\gamma}>0 \] So the third term is: \[ \cot^{-1}(-K)=\pi-\cot^{-1}(K) \] and since $K>0$, \[ \cot^{-1}(K)=\tan^{-1}\left(\frac{1}{K}\right)=\tan^{-1}\left(\frac{\alpha-\gamma}{1+\alpha\gamma}\right) \] Hence third term becomes: \[ \pi-\tan^{-1}\left(\frac{\alpha-\gamma}{1+\alpha\gamma}\right) =\pi+\tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right) \] --- ### 4) Add all three terms So \[ S=\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)+\tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right)+\pi+\tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right) \] Now set: \[ A=\tan^{-1}\alpha,\quad B=\tan^{-1}\beta,\quad C=\tan^{-1}\gamma \] Since $\alpha>\beta>\gamma>0$, we get: \[ 0\beta>\gamma$ but $\gamma<0$, which step in the solution breaks first and why (hint: principal values/sign of the third term)?

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JEE Mains JEE Advance BITSAT
JEE Mains | PYQ