Question

The value of ${\cos }^2\theta -\cos \theta \cos ({60}^o-\theta )+{\cos }^2({60}^o-\theta )$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

Answer: (C)

$\frac{3}{4}$

$co{s}^2\theta -cos\theta cos(60-\theta )+co{s}^2(60-\theta )$

$cos(A-B)=cosA.cosB+sinA.sinB$

$=co{s}^2\theta -cos\theta (cos60.cos\theta +sin60.sin\theta )+(cos60.cos\theta +sin60.sin\theta {)}^2$

$=co{s}^2\theta -cos\theta (\frac{cos\theta }{2}+\frac{\sqrt{3}sin\theta }{2})+(\frac{cos\theta }{2}+\frac{\sqrt{3sin\theta }}{2}{)}^2$

$=co{s}^2\theta -\frac{co{s}^2\theta }{2}-\frac{\sqrt{3}sin\theta .cos\theta }{2}+\frac{co{s}^2\theta }{4}+\frac{2\sqrt{3}sin\theta .cos\theta }{4}+\frac{3si{n}^2\theta }{4}$

$=\frac{3co{s}^2\theta }{4}+\frac{3si{n}^2\theta }{4}$

$=\frac{3}{4}(si{n}^2\theta +co{s}^2\theta )$

$=\frac{3}{4}$