$ \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) $ is equal to
(A) $ 2 \cos \theta $
(B) 0
(C) $ 2 \sin \theta $
(D) 1

Given:

\( \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) \)

To do:

We have to find the value of \( \sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right) \).

Solution:

We know that,

$\cos(90^o - \theta) = \sin \theta$

Therefore,

$\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)= \cos[90^o- (45^o + \theta)] - \cos(45^o- \theta)$

$= \cos (45^o - \theta) - \cos (45^o - \theta)$

$= 0$

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Updated on: 10-Oct-2022

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