Question

The value of ${\sin }^2({90}^{\circ }-\theta )[1+{\cot }^2({90}^{\circ }-\theta )]$ is

• A. -1
• B. 0
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (D)

1

Given ${\sin }^2({90}^{\circ }-\theta )[1+{\cot }^2({90}^{\circ }-\theta )]$ $\dots I$

we know that $\sin ({90}^{\circ }-\theta )=\cos \theta$

$\cot ({90}^{\circ }-\theta )=\tan \theta$

substitute these in $I$ it becomes

$={\cos }^2\theta [1+{\tan }^2\theta ]$

$={\cos }^2\theta \left[{\sec }^2\theta \right]$

$=1$

$\therefore {\sin }^2({90}^{\circ }-\theta )[1+{\cot }^2({90}^{\circ }-\theta )]=1$