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Question

Mathematics Question on Trigonometric Functions

cos⁡θ.cos⁡(90−θ)−sin⁡θsin⁡(90−θ)\cos \theta .\cos(90 - \theta) - \sin \theta \sin (90 - \theta) equals:

A

1

B

2

C

-1

D

0

Answer

0

Explanation

Solution

Consider cos⁡θ.cos⁡(90−θ)−sin⁡ θ sin⁡(90−θ)\cos \theta . \cos(90 - \theta) - \sin \, \theta \, \sin(90 - \theta) =cos⁡θ.sin⁡ θ−sin⁡ θ.cos⁡ θ=0= \cos \theta . \sin \, \theta - \sin \, \theta . \cos \, \theta = 0