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Question: \[\mathbf{\cos}\mathbf{A}\mathbf{+}\mathbf{\sin}\mathbf{(}\mathbf{270}\mathbf{{^\circ}}\mathbf{+ A)}...

cos⁡A+sin⁡(270∘+A)−sin⁡(270∘−A)+cos⁡(180∘+A)=\mathbf{\cos}\mathbf{A}\mathbf{+}\mathbf{\sin}\mathbf{(}\mathbf{270}\mathbf{{^\circ}}\mathbf{+ A)}\mathbf{-}\mathbf{\sin}\mathbf{(}\mathbf{270}\mathbf{{^\circ} -}\mathbf{A) +}\mathbf{\cos}\mathbf{(}\mathbf{180}\mathbf{{^\circ}}\mathbf{+ A) =}

A

– 1

B

0

C

1

D

None of these

Answer

0

Explanation

Solution

cos⁡A+sin⁡(270∘+A)−sin⁡(270∘−A)+cos⁡(180∘+A)\cos A + \sin(270{^\circ} + A) - \sin(270{^\circ} - A) + \cos(180{^\circ} + A)

=cos⁡A−cos⁡A+cos⁡A−cos⁡A=0\cos A - \cos A + \cos A - \cos A = 0.