Question

In a$∆ABC,\text{if}\angle A=\pi /2,then{\cos }^{2}B+{\cos }^{2}C$ equals to

• A. $-2$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

The correct option is C

$1$

Determine the value of : ${\cos }^{2}B+{\cos }^{2}C$

We know that the interior angles of a triangle $180°\text{or}\mathrm{\pi }$.

$$\begin{gathered} \therefore \angle A+\angle B+\angle C=\mathrm{\pi } \\ \Rightarrow \angle \mathrm{B}+\angle \mathrm{C}=\mathrm{\pi }-\frac{\mathrm{\pi }}{2}...\left(1\right)\left[\because \angle \mathrm{A}=\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$

From trigonometric identities we have,

$$\begin{gathered} \Rightarrow {\cos }^{2}B+{\cos }^{2}C={\cos }^{2}B+{\cos }^2\left(\frac{\mathrm{\pi }}{2}-B\right)\left[\because \angle B+\angle C=\frac{\mathrm{\pi }}{2}\right] \\ \Rightarrow ={\cos }^{2}B+{\sin }^{2}B[\because \cos \left(90°-x\right)=\sin (x\left)\right] \\ \Rightarrow {\cos }^{2}B+{\cos }^{2}C=1[\because {\cos }^{2}B+{\sin }^{2}B=1] \end{gathered}$$

Thus, option (C) is correct.