Question 95

If A = 30°, B = 60° and C = 135°, then what is the value of $$sin^3A + cos^3B + tan^3C - 3sin A cos B tan C$$ ?

Solution

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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