Question

cos10°-sin10°=k then cos20°​

Answer 1

Your answer is in the attachment provided.

so please refer it.

Thnks

Answer 2

Trigonometry

$cos10\textdegree-sin10\textdegree=k$

On squaring both sides,

$(cos10\textdegree-sin10\textdegree)^2=k^2\\\\\implies (cos10\textdegree)^2+(sin10\textdegree)^2+2 (cos10\textdegree)(sin10\textdegree)=k^2\\\\\implies cos^210\textdegree+sin^210\textdegree+2 (cos10\textdegree)(sin10\textdegree)=k^2$

$\implies 1+sin2(10\textdegree)=k^2$

Since,

$sin^2\theta+cos^2\theta=1\\\\and\\\\2sin\theta cos\theta=sin2\theta$

$\implies 1-sin20\textdegree=k^2\\\\\implies 1-k^2=sin20\textdegree$

By pythagoras theorem,

$cos20\textdegree=\frac{\sqrt{1-(1-k^2)^2} }{1} \\\\\implies cos 20\textdegree =\sqrt{(k^2)(2-k^2)}\\ \\\implies cos 20\textdegree=k\sqrt{2-k^2}$

Hence, cos 20°= k√(2-k²).