Question

If $\cos {25}^o-\sin {25}^o=k$ then $\cos {50}^o$ is equal to-

• A. $k\sqrt{2-k^2}$
• B. $-\sqrt{2-k^2}$
• C. $\sqrt{2-k^2}$
• D. $-k\sqrt{2-k^2}$

Answer 1

Answer: (A)

$k\sqrt{2-k^2}$

$\cos {50}^o=(\cos {25}^o-\sin {25}^o)(\cos {25}^o+\sin {25}^o)$
We have,
$(\cos {25}^o+\sin {25}^o{)}^2=2-(\cos {25}^o+\sin {25}^o{)}^2$
$\Rightarrow \cos {25}^o+\sin {25}^o=\sqrt{2-k^2}$
$\Rightarrow \cos {50}^o=k\times \sqrt{2-k^2}$