Q64 / 180 · KCET 2009 · Mathematics

$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}$ $+4\left(\sin ^{6} x+\cos ^{6} x\right)$ is equal to

  • A
    12
  • Correct answer
    B
    13
  • C
    14
  • D
    11

Solution

$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}$ $+4\left(\sin ^{6} x+\cos ^{6} x\right)$ $=3(1-2 \sin x \cos x)^{2}+6(1+2 \sin x \cos x)$ $\quad+4\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x\right.$ $\left.\quad-\sin ^{2} x \cos ^{2} x\right)$ $=3\left[1+4 \sin ^{2} x \cos ^{2} x-4 \sin x \cos x\right]$ $\quad+6+12 \sin x \cos x+4\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}\right.$ $\left.\quad-2 \sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x\right]$

Hence, the correct answer is option: 'B'.

Relevant Questions