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)

$0$

Answer 2

_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)