Question

Simplify the following expression \[\dfrac{\sin \left( -\theta \right)\tan \left( 90+\theta \right)\sin \left( 180+\theta \right)\sec \left( 270+\theta \right)}{\sin \left( 360-\theta \right)\cos \left( 270-\theta \right)\text{cosec}\left( 180-\theta \right)\cot \left( 360-\theta \right)}\]

Answer 1

Hint: We have an expression as \[\dfrac{\sin \left( -\theta \right)\tan \left( 90+\theta \right)\sin \left( 180+\theta \right)\sec \left( 270+\theta \right)}{\sin \left( 360-\theta \right)\cos \left( 270-\theta \right)\text{cosec}\left( 180-\theta \right)\cot \left( 360-\theta \right)}\]. Firstly, simplify the compound angles by using the trigonometric identities. Then, simplify the negative of angles. We get a simple expression in terms of $\theta $ . Try to cancel out the terms and get a finite value of expression.

Complete step-by-step solution
We have been given the expression as
\[\Rightarrow \dfrac{\sin \left( -\theta \right)\tan \left( 90+\theta \right)\sin \left( 180+\theta \right)\sec \left( 270+\theta \right)}{\sin \left( 360-\theta \right)\cos \left( 270-\theta \right)\text{cosec}\left( 180-\theta \right)\cot \left( 360-\theta \right)}......(1)\]
First, we will try to convert the complex terms into simpler terms.
We know that we have certain conversions possible. So, the different conversions for compound angles are enlisted below,
\[\begin{align}
  & \tan \left( 90+\theta \right)=-\cot \theta \\
 & \sin \left( 180+\theta \right)=-\sin \theta \\
 & \sec \left( 270+\theta \right)=\text{cosec}\theta \\
 & \sin \left( 360-\theta \right)=-\sin \theta \\
 & \cos \left( 270-\theta \right)=-\sin \theta \\
 & \text{cosec}\left( 180-\theta \right)=\text{cosec}\theta \\
 & \cot \left( 360-\theta \right)=-\cot \theta \\
\end{align}\]
Now, using the above and simplifying all the compound angles in equation (1), we get:
\[\Rightarrow \dfrac{\sin \left( -\theta \right)\times -\cot \theta \times -\sin \theta \times \text{cosec}\theta }{-\sin \theta \times -\sin \theta \times \text{cosec}\theta \times -\cot \theta }......(2)\]
Now, simplifying the sine of negative angle as negative of sine of angle in equation (2), we
get:\[\Rightarrow \dfrac{-\sin \theta \times -\cot \theta \times -\sin \theta \times \text{cosec}\theta }{-\sin \theta \times -\sin \theta \times \text{cosec}\theta \times -\cot \theta }......(3)\]
Now, cancelling out similar terms in equation (3), we get:
\[\Rightarrow \dfrac{-\sin \theta \times -\cot \theta \times -\sin \theta \times \text{cosec}\theta }{-\sin \theta \times -\sin \theta \times \text{cosec}\theta \times -\cot \theta }=1\]
Hence, we have got the value of expression \[\dfrac{\sin \left( -\theta \right)\tan \left( 90+\theta \right)\sin \left( 180+\theta \right)\sec \left( 270+\theta \right)}{\sin \left( 360-\theta \right)\cos \left( 270-\theta \right)\text{cosec}\left( 180-\theta \right)\cot \left( 360-\theta \right)}\] is 1.

Note: Keep in mind that the quadrant system for the simplification of compound angles. In the first quadrant, all the angles are positive. In the second quadrant, sine and cosecant of angles are positive. In the third quadrant, tangent and cotangent of angles are positive, and in the fourth quadrant cosine and secant of angles are positive.
Also, an odd multiple of 90 changes the angle whereas the even multiple remains the same, i.e.
$\sin \left( 90+\theta \right)=\cos \theta $ but $\sin \left( 180+\theta \right)=-\sin \theta $