Q.If $\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ equals
(A) $0$
(B) $1$
(C) $6$
(D) $12$

Concept: Range of inverse cosine and sum of angles.

The principal value of $\cos^{-1}x$ lies in $[0,\pi]$. For three such angles to sum to $3\pi$, each must individually be $\pi$ (the maximum possible).

Thus:

$\cos^{-1}\alpha = \pi \Rightarrow \alpha = -1$

$\cos^{-1}\beta = \pi \Rightarrow \beta = -1$

$\cos^{-1}\gamma = \pi \Rightarrow \gamma = -1$

Now compute: …

Concept & Intuition

The range of $\cos^{-1}x$ is $[0, \pi]$. The maximum possible value of each term is $\pi$, so the sum of three such terms can be at most $3\pi$. The given condition says the sum equals $3\pi$, which is the absolute maximum. This can only happen if every term individually hits its maximum: $\cos^{-1}\alpha = \pi$, $\cos^{-1}\beta = \pi$, $\cos^{-1}\gamma = \pi$.

Why? If any term were less than $\pi$, the total would be strictly less than $3\pi$. So the condition forces each inverse cosine to be exactly $\pi$, meaning each argument is $\cos\pi = -1$.

Now the expression $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ simplifies nicely when all three variables are equal to $-1$.

Step-by-step solution

1. Identify the range constraint For any real $x$ in $[-1,1]$, $\cos^{-1}x \in [0,\pi]$. Therefore:

2. Sum to maximum forces each term to be maximum Adding the inequalities:

The given sum is exactly $3\pi$, the upper bound. This is only possible if each term equals its maximum:

3. Find the values of $\alpha, \beta, \gamma$ Taking cosine on both sides:

4. Compute the required expression The expression is: …

Method: Extremal (boundary-hitting) argument for a sum of bounded inverse functions

When a sum of inverse-trig terms equals the largest (or smallest) value it could possibly take, every term must individually be at its extreme. This pins down each variable exactly.

Steps

Step 1: Bound each term by its range.

Each $\cos^{-1}$ lies in $[0,\pi]$, so a sum of three is at most $3\pi$. Write the inequality:

Step 2: Use the equality condition.

If the given sum equals the upper bound $3\pi$, equality in the sum forces equality in each term: $\cos^{-1}\alpha=\cos^{-1}\beta=\cos^{-1}\gamma=\pi$. (If any term were smaller, the total would fall short.) …

Common Mistakes

Mistake 1: Thinking $\cos^{-1}(-1)=0$ (or $\tfrac{\pi}{2}$).

Why it's wrong: $\cos^{-1}(-1)=\pi$, the top of the $[0,\pi]$ range. Correct approach: for each inverse cosine to equal its maximum $\pi$, the argument must be $\cos\pi=-1$.

Mistake 2: Not realising the sum $3\pi$ forces every term to its maximum. …