In triangle ABC,if cos ^2 A + cos ^2 B + cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$. Using the identity $\cos ^2 C = 1 - \sin ^2 C$,we have: $\cos ^2 A + \cos ^2 B + 1 - \sin ^2 C = 1$

Vedclass · Accepted Answer

a right-angled triangle

Vedclass · Answer

(C) Given $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$. Using the identity $\cos ^2 C = 1 - \sin ^2 C$,we have: $\cos ^2 A + \cos ^2 B + 1 - \sin ^2 C = 1$ $\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2 C$ Since $C = 180^{\circ} - (A + B)$,$\sin C = \sin(A + B)$. $\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2(A + B)$ $\Rightarrow \cos ^2 A + \cos ^2 B = (\sin A \cos B + \cos A \sin B)^2$ $\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2 A \cos ^2 B + \cos ^2 A \sin ^2 B + 2 \sin A \cos B \cos A \sin B$ $\Rightarrow \cos ^2 A(1 - \sin ^2 B) + \cos ^2 B(1 - \sin ^2 A) = 2 \sin A \cos B \cos A \sin B$ $\Rightarrow \cos ^2 A \cos ^2 B + \cos ^2 B \cos ^2 A = 2 \sin A \cos B \cos A \sin B$ $\Rightarrow 2 \cos ^2 A \cos ^2 B - 2 \sin A \cos B \cos A \sin B = 0$ $\Rightarrow 2 \cos A \cos B (\cos A \cos B - \sin A \sin B) = 0$ $\Rightarrow 2 \cos A \cos B \cos(A + B) = 0$ Since $\cos(A + B) = \cos(180^{\circ} - C) = -\cos C$,we get: $-2 \cos A \cos B \cos C = 0$ Thus,$\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$. This implies $A = 90^{\circ}$ or $B = 90^{\circ}$ or $C = 90^{\circ}$. Therefore,$	riangle ABC$ is a right-angled triangle.

In $\triangle ABC$,if $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$,then $\triangle ABC$ is

Similar Questions

The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for

Let $p_1, p_2, p_3$ be the altitudes of a triangle $ABC$ drawn through the vertices $A, B, C$ respectively. If $r_1=4, r_2=6, r_3=12$ are the ex-radii of triangle $ABC$, then $\frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=$

In $\Delta ABC$,the expression $a(\cos^2 B + \cos^2 C) + \cos A(c \cos C + b \cos B)$ is equal to:

In a $\triangle ABC$,if $A-B=120^{\circ}$ and $R=8r$,then find the value of $\frac{1+\cos C}{1-\cos C}$.

In a triangle $ABC$,if $(a-b)(s-c)=(b-c)(s-a)$,then $r_1+r_3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module