(sin 40° − cos 50°) = ?
(a) sin 10°
(b) cos 10°
(c) 1
(d) 0

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Solution

$(d) 0 We have: (\sin 40^{0} - \cos 50^{0}) = \sin 40^{0} - \cos (90^{0} - 40^{0}) = \sin 40^{0} - \sin 40^{0} [∵ \cos (90^{0} - θ) = \sin θ] = 0$

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