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8. Introduction to Trigonometry
easy
$\sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right)$ बराबर है
A
$2 \cos \theta$
B
$0$
C
$2 \sin \theta$
D
$1$
Solution
$\sin \left(45^{\circ}+\theta\right)-\cos \left(45^{\circ}-\theta\right)$
$=\cos \left[90^{\circ}-\left(45^{\circ}+\theta\right)\right]-\cos \left(45^{\circ}-6\right)\left[\therefore \cos \left(90^{\circ}-\theta\right)=\sin 0\right]$
$=\cos \left(45^{\circ}-0\right)-\cos \left(45^{\circ}-0\right)$
$=0$
Standard 10
Mathematics
