SSC CGL 9 Aug 2017 Shift 1 Paper

Question : 70 of 100

Marks: +1, -0

If A =30°,B =60° and C =135°, then what is the value of

\sin^3 A + \cos^3 B + \tan^3 C

-3 \sin A \cos B \tan C

Solution:

A =30°,B =60°, C =135°

\sin^3 A + \cos^3 B + \tan^3 C

-3 \sin A \cos B \tan C

= \sin^3 30° + \cos^3 60° + \tan^3 135°

-3 \sin30° . \cos 60° . \tan 135°

=(1/2)^3+(1/2)^3 -\cot45°

-3×1/2×1/2×(-\cot45°)

1/8+1/8-1+3/4×1=2/8-1+3/4

=-3/4+3/4=0

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