SSC CHSL Model Paper 11

Question : 70 of 100

Marks: +1, -0

\cos x + \cos^2x = 1

, the numerical value of

(\sin^12x + 3 \sin^10x

+ 3 \sin^8x + \sin^6x -1)

is :

Solution:

\cos x + \cos^2 x = 1

⇒\cos x = 1 - \cos^2 x = \sin^2 x

∴\sin^12x + 3 \sin^10x

+ 3 \sin^8x + \sin^6x -1

=(\sin^4x+\sin^2x)^3-1

=(\cos^2x+\sin^2x)-1

=1-1=0

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