Q84 / 200 · WBJEE 2013 · Mathematics

The value of $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}$ $-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ is

  • A
    0
  • B
    1
  • Correct answer
    C
    $\frac{1}{2}$
  • D
    $\frac{1}{4}$

Solution

Given $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}\cos^260^\circ$ $=\left(\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)^{2}$ $-\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$ $=\left(\frac{3+1-2 \sqrt{3}}{8}\right)+\frac{1}{2}+\left(\frac{3+1+2 \sqrt{3}}{8}\right)-\frac{3}{4}-\frac{1}{4}$ $=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{2}+\frac{1}{2}+\frac{\sqrt{3}}{4}-\frac{3}{4}-\frac{1}{4}=\frac{3}{2}-1=\frac{1}{2}$

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

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