Question

In a triangle ABC, if $\cos^2 A - \sin^2 B + \cos^2 C=0$, then the value of $\cos A \cos B \cos C$ is

• A. $\frac{1}{4}$
• B. 1
• C. $\frac{\pi}{2}$
• D. $\frac{1}{2}$
• E. 0

Answer 1

Answer: (E)

0

Answer 2

We are given the triangle $ABC$ with

$$\cos^2 A - \sin^2 B + \cos^2 C = 0.$$

Rewrite $\sin^2 B$ as:

$$\sin^2 B = 1 - \cos^2 B.$$

Substitute:

$$\cos^2 A - (1 - \cos^2 B) + \cos^2 C = 0 \implies \cos^2 A + \cos^2 B + \cos^2 C - 1 = 0.$$

Thus,

$$\cos^2 A + \cos^2 B + \cos^2 C = 1. \tag{1}$$

A known trigonometric identity for a triangle is:

$$\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1. \tag{2}$$

Comparing $(1)$ and $(2)$:

$$1 + 2 \cos A \cos B \cos C = 1 \implies 2 \cos A \cos B \cos C = 0.$$

Thus,

$$\cos A \cos B \cos C = 0.$$