Question

Evaluate the following:
$\frac{\tan {35}^0}{\cot {55}^0}+\frac{\cot {78}^0}{\tan {12}^0}-1$

Answer 1

We have $\tan (90-\theta )=\cot \theta$ and $\cot (90-\theta )=\tan \theta$

therefore,

$\tan 35=\tan ({90}^0-{55}^0)=\cot {55}^0$

$\cot 78=\cot ({90}^0-{78}^0)=\tan {12}^0$

To evaluate,

$\frac{\tan {35}^0}{\cot {55}^0}+\frac{\cot {78}^0}{\tan {12}^0}-1$

$=\frac{\cot {55}^0}{\cot {55}^0}+\frac{\tan {12}^0}{\tan {12}^0}-1$

$=1+1-1$

$=1$