Question

Find the value of $\alpha$ for which the equality $\frac{\cos {10}^o+\sin {10}^o}{\cos {10}^o-\sin {10}^o}=\cot \alpha$ holds?

Answer 1

Given,

$\frac{\cos {10}^o+\sin {10}^o}{\cos {10}^o-\sin {10}^o}=\cot \alpha$.......(1)

Now,

$\frac{\cos {10}^o+\sin {10}^o}{\cos {10}^o-\sin {10}^o}$

$=\frac{\cot {10}^o+1}{\cot {10}^o-1}$ [ Dividing both the numerator and the denominator by $\sin {10}^o$]

$=\frac{\cot {10}^o.\cot {45}^o+1}{\cot {10}^o-\cot {45}^o}$

$=\cot ({45}^o-{10}^o)$

$=\cot {35}^o$.

From (1) we've $\alpha ={35}^o$.