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

Simplify:
cos⁡2(1800−θ)+cos⁡2(2700+θ)sin⁡(1800+θ)sin⁡(−θ)+cos⁡2(2700+θ)sin⁡(1800+θ)\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)}

Explanation

Solution

Hint: Here, we are going to use the trigonometric conversions as cos⁡(1800−θ)=−cos⁡θ\cos \left( {{180}^{0}}-\theta \right)=-\cos \theta , sin⁡(1800+θ)=−sin⁡θ\sin \left( {{180}^{0}}+\theta \right)=-\sin \theta ,cos⁡(2700+θ)=sin⁡θ\cos \left( {{270}^{0}}+\theta \right)=\sin \theta and sin⁡(−θ)=−sin⁡θ\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⁡(1800−θ)=−cos⁡θ\cos \left( {{180}^{0}}-\theta \right)=-\cos \theta
This is because cosine is negative in the second quadrant.
sin⁡(1800+θ)=−sin⁡θ\sin \left( {{180}^{0}}+\theta \right)=-\sin \theta
This is due to sin being negative in the third quadrant.
cos⁡(2700+θ)=sin⁡θ\cos \left( {{270}^{0}}+\theta \right)=\sin \theta
This is because cosine is positive in the fourth quadrant.
sin⁡(−θ)=−sin⁡θ\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:
cos⁡2(1800−θ)+cos⁡2(2700+θ)sin⁡(1800+θ)sin⁡(−θ)+cos⁡2(2700+θ)sin⁡(1800+θ) ⇒cos⁡2θ+sin⁡2θ−sin⁡θ−sin⁡θ+sin⁡2θ−sin⁡θ \begin{aligned} & \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{aligned}
Now, in the above step we will take −1sin⁡θ-\dfrac{1}{\sin \theta }as common and in the expression of ⇒sin⁡2θ−sin⁡θ\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:
⇒−1sin⁡θ(cos⁡2θ−sin⁡θ+sin⁡2θ)\Rightarrow -\dfrac{1}{\sin \theta }\left( {{\cos }^{2}}\theta -\sin \theta +{{\sin }^{2}}\theta \right)
As we know that the trigonometric identity cos⁡2θ+sin⁡2θ{{\cos }^{2}}\theta+{{\sin }^{2}}\theta = 1. Substituting this identity value in the above expression we get,
⇒−1sin⁡θ(1−sin⁡θ)  \begin{aligned} & \Rightarrow -\dfrac{1}{\sin \theta }\left( 1-\sin \theta \right) \\\ & \\\ \end{aligned}
Now, we are going to open the bracket and the expression will look like as follows:
⇒−1sin⁡θ+sin⁡θsin⁡θ  \begin{aligned} &\Rightarrow -\dfrac{1}{\sin \theta }+\dfrac{\sin \theta }{\sin \theta } \\\ & \\\ \end{aligned}
We know that 1sin⁡θ=cos⁡ecθ\dfrac{1}{\sin \theta }=\cos ec\theta , sin⁡θsin⁡θ=1\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(1800−θ){{\cos }^{2}}\left( {{180}^{0}}-\theta \right) bracket as it look like cos⁡(A−B)\cos \left( A-B \right) identity then you will open the cos⁡2(1800−θ){{\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.