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: In the given question, we only have sine and cosine, which follow the below relation \[\sin \left( {90^\circ - \theta } \right) = \cos \theta \] and hence it’s also clear that this relation is also followed with it\[\cos \left( {90^\circ - \theta } \right) = \sin \theta \], and also keep in mind the basic trigonometric square identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]

Formula Used:
We must be remembering the following basic identities:
\[\sin \left( {90^\circ - \theta } \right) = \cos \theta \]
\[\cos \left( {90^\circ - \theta } \right) = \sin \theta \]
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
We know that,
\[\cos \left( {180^\circ + \theta } \right) = - \cos \theta {\rm{ }}...(i)\]
Also,
\[\sin \left( {180^\circ + \theta } \right) = - \sin \theta {\rm{ }}...\left( {ii} \right)\]
And,
\[\cos \left( {90^\circ + \theta } \right) = - \sin \theta {\rm{ }}...\left( {iii} \right)\]\[\]

Complete step-by-step answer:
Now, we shall approach by solving the first trigonometric identity containing cosine:
\[\cos \left( {270^\circ + \theta } \right)\]
\[ = \cos \left( {180^\circ + 90^\circ + \theta } \right)\]
\[ = \cos \left( {180^\circ + \left( {90^\circ + \theta } \right)} \right)\]
\[ = - \cos \left( {90^\circ + \theta } \right)\] (from (i))
\[ = \sin \theta \] (from (iii))
Now, the other part of the given expression is pretty clear from the basic identities:
\[\cos \left( {90^\circ - \theta } \right) = \sin \theta \]
Thus, \[\cos \left( {270^\circ + \theta } \right)\]\[ \times \]\[\cos \left( {90^\circ - \theta } \right)\]\[ = \]\[{\sin ^2}\theta {\rm{ }}...\left( {iv} \right)\]
Now we shall solve the given expression after the minus sign.
It contains one straight-forward expression and one not so straight-forward expression but we will get by, so we have:
\[\sin \left( {270^\circ - \theta } \right) \times \cos \theta \]
Now, \[\sin \left( {270^\circ - \theta } \right)\]
\[ = \sin \left( {180^\circ + (90^\circ - \theta )} \right)\]
\[ = - \sin \left( {90^\circ - \theta } \right)\] (from (ii))
But by applying the basic identities we get,
\[ = - \cos \theta \]
So,
\[\sin \left( {270^\circ - \theta } \right) \times \cos \theta = \]\[ - \cos \theta \times \cos \theta = \]\[ - {\cos ^2}\theta \]
or, \[ - \sin \left( {270^\circ - \theta } \right) \times \cos \theta = \]\[{\rm{co}}{{\rm{s}}^2}\theta {\rm{ }}...\left( v \right)\]
Finally adding the two simplified equations (iv) and (v) to get the answer of the original question, we have,
\[\cos \left( {270^\circ + \theta } \right)\cos \left( {90^\circ - \theta } \right) - \sin \left( {270^\circ - \theta } \right)\cos \theta \]
\[ = {\sin ^2}\theta + {\cos ^2}\theta = 1\]
Hence, the answer of the question is d) 1.

Note: In questions like these, we have to give extra care to the signs (negative/positive) before the trigonometric ratios. Then, students must remember the periodicity of the different trigonometric functions, when they change their signs, what difference occurs when the argument of the function is \[180 + \theta ,180 - \theta ,270 + \theta ,\] etc. They do not always come right away. You need to have all your bases clear, remember the basic identities (because that is where the final thing usually comes down to) and then accordingly mold the equation in the required form. We must just be careful, pay attention and solve the question in the flow and it will always work out.