Question

$\frac{\sec \theta }{\text{cosec}({90}^o-\theta )}-\frac{\sin \theta }{\cos ({90}^o-\theta )}+\cos 0^o$ is equal to :

• A. $1$
• B. $3$
• C. $2$
• D. $0$

Answer 1

Answer: (A)

$1$

We need to find value of $\frac{\sec \theta }{\text{cosec}({90}^o-\theta )}-\frac{\sin \theta }{\cos ({90}^o-\theta )}+\cos 0^o$

$=\frac{\sec \theta }{\sec \theta }-\frac{\sin \theta }{\sin \theta }+1$

$=1-1+1$

$=1$