sin (60° + θ) – cos (30° – θ) = ?

Question:

sin (60° + θ) – cos (30° – θ) = ?
(a) 2sin θ
(b) 2cos θ
(c) 0
(d) 1

 

Solution:

$\sin \left(60^{\circ}+\theta\right)-\cos \left(30^{\circ}-\theta\right)$

$=\cos \left(90^{\circ}-\left(60^{\circ}+\theta\right)\right)-\cos \left(30^{\circ}-\theta\right) \quad\left(\because \sin \theta=\cos \left(90^{\circ}-\theta\right)\right)$

$=\cos \left(90^{\circ}-60^{\circ}-\theta\right)-\cos \left(30^{\circ}-\theta\right)$

$=\cos \left(30^{\circ}-\theta\right)-\cos \left(30^{\circ}-\theta\right)$

$=0$

Hence, the correct option is (c).

 

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