Question

If ${\cos ^2}A + co{s^2}B + {\cos ^2}C = 1,$ then $\Delta ABC$ is
A) Equilateral
B) Isosceles
C) Scalene
D) Right angled

Answer 1

Hint:
Given ${\cos ^2}A + co{s^2}B + {\cos ^2}C = 1,$ we know that $\cos 2\theta = 2{\cos ^2}\theta - 1$ then convert ${\cos ^2}A$ and $co{s^2}B$ into $\cos 2A$ and $\cos 2B$. Then equation will be $\cos 2A + \cos 2B + 2{\cos ^2}C = 0$. we will use $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ for simplification. After simplification we will get $\cos C\cos B\cos A = 0$ this is possible only when any one angle is the right angle triangle.

Complete step by step solution:
Given ${\cos ^2}A + co{s^2}B + {\cos ^2}C = 1,$
Multiplying 2 to both side
$2{\cos ^2}A + 2co{s^2}B + 2{\cos ^2}C = 2$
Convert ${\cos ^2}A$ and $co{s^2}B$ into $\cos 2A$ and $\cos 2B$ by using $\cos 2\theta = 2{\cos ^2}\theta - 1$
$\cos 2A + 1 + \cos 2B + 1 + 2{\cos ^2}C = 2$
$\cos 2A + \cos 2B + 2{\cos ^2}C = 0$
It is known that $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ then equation will be
$2\cos \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) + 2{\cos ^2}C = 0$
$2\cos \left( {A + B} \right)\cos \left( {A - B} \right) + 2{\cos ^2}C = 0$
ABC is a triangle sum of all the angle is ${180^ \circ }$ so $A + B = \pi - C$ and $cos\left( {\pi - \theta } \right) = - \cos \theta $
$ - 2\cos C\cos \left( {A - B} \right) + 2{\cos ^2}C = 0$
Taking common $2\cos C$
$ - 2\cos C\left( {\cos \left( {A - B} \right) - \cos C} \right) = 0$
Again using $\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)$ we get
$\cos C\sin \left( {\dfrac{{A - B + C}}{2}} \right)\sin \left( {\dfrac{{A - B - C}}{2}} \right) = 0$
$A + C = \pi - B$ and $B + C = \pi - A$
$\cos C\sin \left( {\dfrac{\pi }{2} - B} \right)\sin \left( {A - \dfrac{\pi }{2}} \right) = 0$
We know that $\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta $ then
$\cos C\cos B\cos A = 0$
This can be 0 only if and only if one angle of the triangle is the right angle.

Hence proved ABC is a right angled triangle.

Note:
Formula used in this question are
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)$
$\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta $