Question

Simplify:
$\dfrac{{{\cos }^{2}}\left( {{180}^{0}}-\theta \right)+\dfrac{{{\cos }^{2}}\left( {{270}^{0}}+\theta \right)}{\sin \left( {{180}^{0}}+\theta \right)}}{\sin \left( -\theta \right)}+\dfrac{{{\cos }^{2}}\left( {{270}^{0}}+\theta \right)}{\sin \left( {{180}^{0}}+\theta \right)}$

Answer 1

Hint: Here, we are going to use the trigonometric conversions as $\cos \left( {{180}^{0}}-\theta \right)=-\cos \theta $, $\sin \left( {{180}^{0}}+\theta \right)=-\sin \theta $,$\cos \left( {{270}^{0}}+\theta \right)=\sin \theta $and $\sin \left( -\theta \right)=-\sin \theta $. And then substitute these conversions in the trigonometric expression given in the question.

Complete step-by-step answer:
The trigonometric conversions that we are going to use in solving the expression given in the question are as follows:
$\cos \left( {{180}^{0}}-\theta \right)=-\cos \theta $
This is because cosine is negative in the second quadrant.
$\sin \left( {{180}^{0}}+\theta \right)=-\sin \theta $
This is due to sin being negative in the third quadrant.
$\cos \left( {{270}^{0}}+\theta \right)=\sin \theta $
This is because cosine is positive in the fourth quadrant.
$\sin \left( -\theta \right)=-\sin \theta $
This is because sin is negative in the fourth quadrant.
Now, substituting these trigonometric conversions in the expression given in the question we get:
$\begin{align}
  & \dfrac{{{\cos }^{2}}\left( {{180}^{0}}-\theta \right)+\dfrac{{{\cos }^{2}}\left( {{270}^{0}}+\theta \right)}{\sin \left( {{180}^{0}}+\theta \right)}}{\sin \left( -\theta \right)}+\dfrac{{{\cos }^{2}}\left( {{270}^{0}}+\theta \right)}{\sin \left( {{180}^{0}}+\theta \right)} \\
 & \Rightarrow \dfrac{{{\cos }^{2}}\theta +\dfrac{{{\sin }^{2}}\theta }{-\sin \theta }}{-\sin \theta }+\dfrac{{{\sin }^{2}}\theta }{-\sin \theta } \\
\end{align}$
Now, in the above step we will take $-\dfrac{1}{\sin \theta }$as common and in the expression of $\Rightarrow \dfrac{{{\sin }^{2}}\theta }{-\sin \theta }$which lies in the numerator of the first rational expression; sin θ will be cancelled out from numerator and denominator and what will remain look like:
$\Rightarrow -\dfrac{1}{\sin \theta }\left( {{\cos }^{2}}\theta -\sin \theta +{{\sin }^{2}}\theta \right)$
As we know that the trigonometric identity ${{\cos }^{2}}\theta+{{\sin }^{2}}\theta$ = 1. Substituting this identity value in the above expression we get,
$\begin{align}
  & \Rightarrow -\dfrac{1}{\sin \theta }\left( 1-\sin \theta \right) \\
 & \\
\end{align}$
Now, we are going to open the bracket and the expression will look like as follows:
$\begin{align}
  &\Rightarrow -\dfrac{1}{\sin \theta }+\dfrac{\sin \theta }{\sin \theta } \\
 & \\
\end{align}$
We know that $\dfrac{1}{\sin \theta }=\cos ec\theta $, $\dfrac{\sin \theta }{\sin \theta }=1$, substituting these values in the above expression we get,
$\Rightarrow $ -cosec θ + 1
So, the expression given in the question simplifies to –cosec 𝛉 + 1.
Hence, the answer is –cosec θ + 1.

Note: We might think of opening the ${{\cos }^{2}}\left( {{180}^{0}}-\theta \right)$ bracket as it look like $\cos \left( A-B \right)$ identity then you will open the ${{\cos }^{2}}\left( {{180}^{0}}-\theta \right)$ using this identity. The approach is correct but it’s a lengthy process as compared to the one that I have used in the above solution.