1. If A = 30°, B = 60° and C = 135°, then what is the value of $$sin^3A + cos^3B + tan^3C - 3sin A cos B tan C$$ ?
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By: anil on 05 May 2019 01.48 am
Given : A = 30°, B = 60° and C = 135° To find : $$sin^3A + cos^3B + tan^3C - 3sin A cos B tan C$$ = $$[sin(30^circ)]^3+[cos(60^circ)]^3+[tan(135^circ)]^3-3[sin(30^circ)][cos(60^circ)][tan(90^circ)]$$ = $$(frac{1}{2})^3+(frac{1}{2})^3+(-1)^3-3(frac{1}{2})(frac{1}{2})(-1)$$ = $$frac{1}{8}+frac{1}{8}-1+frac{3}{4}$$ = $$frac{1}{4}-frac{1}{4}=0$$ => Ans - (A)
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