Question

If $A+B+C={180}^o$ then the value of $secA(cosBcosC-sinBsinC)$ is?

• A. $-2$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

Answer: (B)

$-1$

We can say,

$secA(cosBcosC-sinBsinC)=secA\left(cos\right(B+C\left)\right)$

($\because$ $cos(B+C)=cosBcosC-sinBsinC$)

$=secA\times \left(cos\right({180}^o-A\left)\right)$

($\because$ $A+B+C={180}^o$, $B+C={180}^o-A$)

$=secA(-cosA)$ ($\because$ $cos({180}^o-A)=-cosA$)

$=-1$ ($\because$ $secA=\frac{1}{cosA}$)