Question

The value of $\sin \theta \cos \theta -\frac{\sin \theta \cos ({90}^o-\theta )\cos \theta }{\sec ({90}^o-\theta )}-\frac{\cos \theta \sin ({90}^o-\theta )\sin \theta }{\text{cosec}({90}^o-\theta )}$ is :

• A. $-1$
• B. $2$
• C. $-2$
• D. $0$

Answer 1

The correct option is D $0$

Let $A=\sin \theta \cos \theta -\frac{\sin \theta \cos ({90}^o-\theta )\cos \theta }{\sec ({90}^o-\theta )}-\frac{\cos \theta \sin ({90}^o-\theta )\sin \theta }{\text{cosec}({90}^o-\theta )}$
Using the identities $\cos ({90}^o-\theta )=\sin \theta ,\sin ({90}^o-\theta )=\cos \theta ,\sec ({90}^o-\theta )=\text{cosec}\theta ,\text{cosec}({90}^o-\theta )=\sec \theta$, we get

$A=\sin \theta \cos \theta -\frac{\sin \theta \sin \theta \cos \theta }{\text{cosec}\theta }-\frac{\cos \theta \cos \theta \sin \theta }{\sec \theta }$

$=\sin \theta \cos \theta -{\sin }^3\theta \cos \theta -{cos}^3\theta \sin \theta$ ..... $[\because \frac{1}{\text{cosec}\theta }=\sin \theta ,\frac{1}{\sec \theta }=\cos \theta ]$

$=\sin \theta \cos \theta -\sin \theta \cos \theta ({\sin }^2\theta +{cos}^2\theta )$

$=\sin \theta \cos \theta -\sin \theta \cos \theta \left(1\right)$

$=\sin \theta \cos \theta -\sin \theta \cos \theta =0$