Question

In a triangle ABC,if \(cos^{2}A-sin^{2}B+cos^{2}C=0\) ,then the value of \(cosAcosBcosC\) is 

• A. \(\dfrac{1}{4}\)
• B. \(1\)
• C. \(\dfrac{\pi}{2}\)
• D. \(\dfrac{1}{2}\)
• E. \(0\)

Answer 1

Answer: (E)

Given that 

In a triangle ABC, 

 \(cos^{2}A-sin^{2}B+cos^{2}C=0\)-----------(1)

then the value of \(cosA.cosB.cosC\) is to be determine

[Here , A , B , C are the angles of triangle ABC]

Case 1

Take, \(A=B=C = 60°\)

Then , test it with equation (1)

We find , 

\(cos^{2}(60°)-sin^{2}(60°)+cos^{2}(60°)\)

\(⇒\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{4} ≠ 0\)

hence the triangle is not an equilateral triangle.

Now take,

 Case-2

Take, \(B =90°\)

then , as we know 

in a triangle \(A+B+C=180°\)

\(⇒C=180°-90°-A\)

\(⇒C=90°-A\)

And \(cos(90°-A)=sinA\)

Now, we can apply it in equation 1 to test as it satisfy the same or not

\(cos^{2}(A)-sin^{2}(90°)+cos^{2}(C)\)

\(=cos^{2}(A)-sin^{2}(90°)+sin^{2}(A)\)

\(=cos^{2}(A)+sin^{2}(A)-sin^{2}(90°)\)

=0 (→ Satisfies the condition)

hence the triangle is not an isosceles triangle.

So,

 \(cosA.cosB.cosC\)

\(=cosA.cos(90°).cosC\)

=\(0\) (_Ans)