Question

The value of $\cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta $ is
A) 0
B) -1
C) $\dfrac{1}{2}$
D) 1

Answer 1

Hint:
Here we will use the periodic identities to simplify the complex terms in the given expression. We will simplify the terms: $\cos \left( {{270}^{\circ }}+\theta \right)$, $\cos \left( {{90}^{{}^\circ }}-\theta \right)$ and $\sin \left( {{270}^{{}^\circ }}-\theta \right)$ using the periodic identities and then we will substitute the simplified value in the given expression. From there, we will get the value of the given expression.

Complete step by step solution:
Given expression is $\cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta $, we need to find the value of the given expression.
We will use periodic identities here.
First we will simplify the term $\cos \left( {{270}^{\circ }}+\theta \right)$ using the periodic identities. We can also write this term as $\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)$. We know from periodic identities that $\cos \left( {{180}^{\circ }}+A \right)=-\cos A$.
Therefore,
$\Rightarrow \cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)=-\cos \left( {{90}^{\circ }}+\theta \right)$ ……. $\left( 1 \right)$
 Now, we will further simplify it. We know from periodic identities that $\cos \left( {{90}^{\circ }}+A \right)=-\sin A$.
Using this periodic identity in equation 1, we get
$\Rightarrow \cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)=-\left( -\sin \theta \right)=\sin \theta $
Hence,
$\Rightarrow \cos \left( {{270}^{\circ }}+\theta \right)=\sin \theta $ …….. $\left( 2 \right)$
Now, we will simplify the term $\sin \left( {{270}^{\circ }}-\theta \right)$ using the periodic identities. We can also write this term as $\sin \left( {{180}^{\circ }}+\left( {{90}^{\circ }}-\theta \right) \right)$. We know from periodic identities that $\sin \left( {{180}^{\circ }}+A \right)=-\sin A$.
Therefore,
$\Rightarrow \sin \left( {{180}^{\circ }}+\left( {{90}^{\circ }}-\theta \right) \right)=-\sin \left( {{90}^{\circ }}-\theta \right)$ ……. $\left( 3 \right)$
 Now, we will further simplify it. We know from periodic identities that $\sin \left( {{90}^{\circ }}-A \right)=\cos A$.
Using this periodic identity in equation 3, we get
$\Rightarrow \sin \left( {{180}^{\circ }}+\left( {{90}^{\circ }}-\theta \right) \right)=-\cos \theta $
Hence,
$\Rightarrow \sin \left( {{270}^{\circ }}-\theta \right)=-\cos \theta $ …….. $\left( 4 \right)$
We know from periodic identities that $\cos \left( {{90}^{\circ }}-A \right)=\sin A$
Therefore, we have
$\Rightarrow \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ ……. $\left( 5 \right)$
Substituting the simplified value of the terms $\cos \left( {{270}^{\circ }}+\theta \right)$, $\cos \left( {{90}^{{}^\circ }}-\theta \right)$ and $\sin \left( {{270}^{{}^\circ }}-\theta \right)$, we get
$\begin{align}
  & \Rightarrow \cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta = \\
 & \sin \theta .\sin \theta -\left( -\cos \theta \right).\cos \theta \\
\end{align}$
Simplifying the terms, we get
$\begin{align}
  & \Rightarrow \cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta = \\
 & \sin \theta .\sin \theta +\cos \theta .\cos \theta \\
\end{align}$
Multiplying the terms, we get
$\begin{align}
  & \Rightarrow \cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta = \\
 & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \\
\end{align}$
We know from trigonometric identities that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.
Therefore,
$\Rightarrow \cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Hence, the required value of $\cos \left( {{270}^{\circ }}+\theta \right)\cos \left( {{90}^{\circ }}-\theta \right)-\sin \left( {{270}^{\circ }}-\theta \right)\cos \theta $ is 1.

Therefore, the correct option is option $\left( d \right)$.

Note:
Periodic identities of trigonometry are also known as co-function identities. Remember that all trigonometric identities are periodic in nature i.e. they repeat their values after a certain period but different trigonometric identities have different periodic constants.