Question

The value of $co{s}^2{10}^o-cos{10}^{o}cos{50}^o+co{s}^2{10}^o$ is equal to

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

Answer 1

The correct option is C $\frac{3}{4}$

Given, $co{s}^2{10}^o-cos{10}^{o}cos{50}^o+co{s}^2{10}^o$

$=\frac{1}{2}(2co{s}^2{10}^o-2cos{10}^{o}cos{50}^o+2co{s}^2{50}^o)$
$=\frac{1}{2}(1+cos{20}^o-(cos{60}^o+cos{40}^o)+1+cos{100}^o)$
$=\frac{1}{2}(2+cos{20}^o-\frac{1}{2}-cos{40}^o+cos({180}^o-{80}^o\left)\right)$
$=\frac{1}{2}(\frac{3}{2}+cos{20}^o-(cos{80}^o+cos{40}^o\left)\right)$
$=\frac{1}{2}(\frac{3}{2}+cos{20}^o-(2cos{60}^{o}cos{20}^o\left)\right)$
$=\frac{1}{2}(\frac{3}{2}+cos{20}^o-(2\left(\frac{1}{2}\right)cos{20}^o)$
$=\frac{1}{2}(\frac{3}{2}+cos{20}^o-cos{20}^o)$
$=\frac{1}{2}\left(\frac{3}{2}\right)$
$=\frac{3}{4}$