Question

Find the value of \[\cos ({{270}^{\circ }}+\theta )\cdot \cos ({{90}^{\circ }}-\theta )-\sin ({{270}^{\circ }}-\theta )\cdot \cos \theta \].
A) 0
B) -1
C) \[-\dfrac{1}{2}\]
 D) 1

Answer 1

Hint: The identities and relations that might be useful in this question are
1) If \[n({{180}^{\circ }})\] is added to an angle or angle is subtracted from \[n({{180}^{\circ }})\], then sin function remains as sin function and cos function remains same as cos function.
2) If \[n({{90}^{\circ }})\] is added to an angle or angle is subtracted from \[n({{90}^{\circ }})\], then sin function gets converted to cos function and cos function gets converted into sin function.
3) The sign of sin and cos should be taken care of while conversion.
4) \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Complete step-by-step answer:
As mentioned in the question, we have to evaluate the given expression. For this, first we will first convert all the angles into the most basic form by using the above property so that only \[\theta \] remains as the angle of input into sin and cos functions.
Now, on simplifying, we can make the expression as follows
\[\begin{align}
  & =\sin (\theta )\cdot \sin (\theta )-(-\cos (\theta ))\cdot \cos \theta \\
\end{align}\] (as sin remains positive in first and second quadrant and cos remains positive in first and fourth quadrant)
\[\begin{align}
 & ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta
    = 1 \\
\end{align}\]
Now, we will use the property mentioned in the hint, we get
The value of the expression will come out as 1.

Note: The students can make an error in finding the value of this expression if the students don’t know the properties which are mentioned in the hint as the question might be trickier otherwise.