Question

Prove that $\frac{\cos 3^o+\sin 3^o}{\cos 3^o-\sin 3^o}=\cot {42}^o$

Answer 1

Now,

$\frac{\cos 3^o+\sin 3^o}{\cos 3^o-\sin 3^o}$

$=\frac{\cot 3^o+1}{\cot 3^o-1}$ [ Dividing the numerator and denominator by $\sin 3^o$ we get]

$=\frac{\cot {45}^o.\cot 3^o+1}{\cot 3^o-\cot {45}^o}$

$=\cot ({45}^o-3^o)$

$=\cot {42}^o$.