Question

Solve:
$\dfrac{\sin \left( {{180}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}+\theta \right)\tan \left( {{270}^{\circ }}-\theta \right)\cot \left( {{360}^{\circ }}+\theta \right)}{\sin \left( {{360}^{\circ }}+\theta \right)\cos \left( {{360}^{\circ }}+\theta \right)\csc \left( -\theta \right)\sin \left( {{270}^{\circ }}+\theta \right)}$
(a) $\sqrt{2}$
(b) $\dfrac{\sqrt{3}}{2}$
(c) $\dfrac{3}{2}$
(d) 1

Answer 1

Hint: To solve the above expression, we are going to use the following trigonometric angle conversions:
$\sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta $, $\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta $, $\tan \left( {{270}^{\circ }}-\theta \right)=\cot \theta $, $\cot \left( {{360}^{\circ }}+\theta \right)=\cot \theta $, $\sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta $, $\cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta $, $\csc \left( -\theta \right)=-\csc \theta $ and $\sin \left( {{270}^{\circ }}+\theta \right)=-\cos \theta $. Now, substitute these conversions in the given trigonometric expression and then solve that expression. While solving that expression, we need to use the following trigonometric properties also: $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$. Hence, this is how we are going to solve the given expression.

Complete step-by-step answer:
The trigonometric expression given in the above problem which we are going to solve is as follows:
$\dfrac{\sin \left( {{180}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}+\theta \right)\tan \left( {{270}^{\circ }}-\theta \right)\cot \left( {{360}^{\circ }}+\theta \right)}{\sin \left( {{360}^{\circ }}+\theta \right)\cos \left( {{360}^{\circ }}+\theta \right)\csc \left( -\theta \right)\sin \left( {{270}^{\circ }}+\theta \right)}$
To solve the above trigonometric expressions we need to have knowledge of trigonometric angle conversions which we are going to show in the below:
$\begin{align}
  & \sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta ; \\
 & \cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta ; \\
 & \tan \left( {{270}^{\circ }}-\theta \right)=\cot \theta ; \\
 & \cot \left( {{360}^{\circ }}+\theta \right)=\cot \theta ; \\
 & \sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta ; \\
 & \cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta ; \\
 & \csc \left( -\theta \right)=-\csc \theta ; \\
 & \sin \left( {{270}^{\circ }}+\theta \right)=-\cos \theta \\
\end{align}$
Now, we are going to substitute the above trigonometric angle conversions in the given expression and we get,
$\dfrac{-\sin \theta \left( -\sin \theta \right)\cot \theta \cot \theta }{\sin \theta \cos \theta \left( -\csc \theta \right)\left( -\cos \theta \right)}$
We know that when we multiply two negative signs then we get positive sign so the two negative signs multiplying in the numerator and the denominator we get,
$\dfrac{\sin \theta \left( \sin \theta \right)\cot \theta \cot \theta }{\sin \theta \cos \theta \left( \csc \theta \right)\left( \cos \theta \right)}$ …………. (1)
Now, we are going to use the following trigonometric conversions of the trigonometric ratios.
$\begin{align}
  & \cot \theta =\dfrac{\cos \theta }{\sin \theta }; \\
 & \csc \theta =\dfrac{1}{\sin \theta } \\
\end{align}$
Using the above trigonometric relations in (1) we get,
$\dfrac{\sin \theta \left( \sin \theta \right)\left( \dfrac{\cos \theta }{\sin \theta } \right)\left( \dfrac{\cos \theta }{\sin \theta } \right)}{\sin \theta \cos \theta \left( \dfrac{1}{\sin \theta } \right)\left( \cos \theta \right)}$
In the numerator of the above expression, $\sin \theta \left( \sin \theta \right)$ will be cancelled out and in the denominator of the above expression, $\sin \theta $ will get cancelled out. Then the above expression will look like:
$\dfrac{\cos \theta \left( \cos \theta \right)}{\cos \theta \left( \cos \theta \right)}$
Now, numerator and denominator is same so both these terms written in the numerator and the denominator will get cancelled out and we get,
1
Hence, the solution of the given expression is 1 and the correct option is 1.

So, the correct answer is “Option (a)”.

Note: The trick to remember the trigonometric conversions which we have done above is that in the first quadrant all $\sin ,\cos ,\tan ,\csc ,\sec ,\cot $ are positive. In the second quadrant, $\sin ,\csc $ is positive, in the third quadrant $\tan ,\cot $ is positive and in the fourth quadrant $\cos ,\sec $.