Question

$\frac{\sin A\cos B+cosecC\cot B+\tan C{\cos }^{2}B}{\sin A\sin C}=$



[1 mark]

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is B 0

$\cos B=\cos 90°=0\text{and}\cot B=\cot 90°=0$

$\frac{\sin A\cos B+cosecC\cot B+\tan C{\cos }^{2}B}{\sin A\sin C}=\frac{\left(\sin A\right)\left(0\right)+\left(cosecB\right)\left(0\right)+\left(\tan C\right)\left(0\right)}{\sin A\sin C}=0$