Question

Find the value of $\sin (\pi +\theta )\sin (\pi -\theta )\cos e{c}^2\theta$.

• A. $1$
• B. $-1$
• C. $\sin \theta$
• D. $-\sin \theta$

Answer 1

The correct option is B

$-1$

Find the value of the given expression

Given expression: $\sin (\pi +\theta )\sin (\pi -\theta )\cos e{c}^2\theta$.

Since $$\sin$$ is a periodic function of time period $$2\pi$$, also positive in second quadrant and negative in third quadrant or $\sin (\pi -\theta )=\sin \theta ,\sin (\pi +\theta )=-\sin \theta$,where$\pi ={180}^{\circ }$

So,

$$\begin{gathered} \sin (\pi +\theta )\sin (\pi -\theta )\cos e{c}^2\theta =-\sin \theta \left(\sin \theta \right)\cos e{c}^2\theta \\ =-{\sin }^2\theta \times \frac{1}{{\sin }^2\theta }[\because cosec\theta =\frac{1}{\mathrm{sin\theta }}] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=-1 \end{gathered}$$

Hence, the value of $\sin (\pi +\theta )\sin (\pi -\theta )\cos e{c}^2\theta$ is $-1$.