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Question: The value of \[\cos \left( {270^\circ + \theta } \right)\cos \left( {90^\circ - \theta } \right) - \...

The value of cos⁡(270∘+θ)cos⁡(90∘−θ)−sin⁡(270∘−θ)cos⁡θ\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.12\dfrac{1}{2}
D. 1

Explanation

Solution

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

Formula Used:
We must be remembering the following basic identities:
sin⁡(90∘−θ)=cos⁡θ\sin \left( {90^\circ - \theta } \right) = \cos \theta
cos⁡(90∘−θ)=sin⁡θ\cos \left( {90^\circ - \theta } \right) = \sin \theta
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
We know that,
cos⁡(180∘+θ)=−cos⁡θ...(i)\cos \left( {180^\circ + \theta } \right) = - \cos \theta {\rm{ }}...(i)
Also,
sin⁡(180∘+θ)=−sin⁡θ...(ii)\sin \left( {180^\circ + \theta } \right) = - \sin \theta {\rm{ }}...\left( {ii} \right)
And,
cos⁡(90∘+θ)=−sin⁡θ...(iii)\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⁡(270∘+θ)\cos \left( {270^\circ + \theta } \right)
=cos⁡(180∘+90∘+θ)= \cos \left( {180^\circ + 90^\circ + \theta } \right)
=cos⁡(180∘+(90∘+θ))= \cos \left( {180^\circ + \left( {90^\circ + \theta } \right)} \right)
=−cos⁡(90∘+θ)= - \cos \left( {90^\circ + \theta } \right) (from (i))
=sin⁡θ= \sin \theta (from (iii))
Now, the other part of the given expression is pretty clear from the basic identities:
cos⁡(90∘−θ)=sin⁡θ\cos \left( {90^\circ - \theta } \right) = \sin \theta
Thus, \cos \left( {270^\circ + \theta } \right)$$$$ \times $$$$\cos \left( {90^\circ - \theta } \right)$$$$ = $$$${\sin ^2}\theta {\rm{ }}...\left( {iv} \right)
Now we shall solve the given expression after the minus sign.
It contains one straight-forward expression and one not so straight-forward expression but we will get by, so we have:
sin⁡(270∘−θ)×cos⁡θ\sin \left( {270^\circ - \theta } \right) \times \cos \theta
Now, sin⁡(270∘−θ)\sin \left( {270^\circ - \theta } \right)
=sin⁡(180∘+(90∘−θ))= \sin \left( {180^\circ + (90^\circ - \theta )} \right)
=−sin⁡(90∘−θ)= - \sin \left( {90^\circ - \theta } \right) (from (ii))
But by applying the basic identities we get,
=−cos⁡θ= - \cos \theta
So,
\sin \left( {270^\circ - \theta } \right) \times \cos \theta = $$$$ - \cos \theta \times \cos \theta = $$$$ - {\cos ^2}\theta
or, - \sin \left( {270^\circ - \theta } \right) \times \cos \theta = $$$${\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⁡(270∘+θ)cos⁡(90∘−θ)−sin⁡(270∘−θ)cos⁡θ\cos \left( {270^\circ + \theta } \right)\cos \left( {90^\circ - \theta } \right) - \sin \left( {270^\circ - \theta } \right)\cos \theta
=sin⁡2θ+cos⁡2θ=1= {\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+θ,180−θ,270+θ,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.