Question

Express the following in terms of trigonometric ratios of angles lying between $0$ and $45$ :
$\tan {65}^0+\cot {49}^0$

Answer 1

We have,

$\tan ({90}^0-\theta )=\cot \theta$ and $\cot ({90}^0-\theta )=\tan \theta$

$\tan {65}^0=\cot ({90}^0-{25}^0)=\cot {25}^0$

$\cot {49}^0=\tan ({90}^0-{41}^0)=\tan {41}^0$

$\therefore \tan {65}^0+\cot {49}^0=\cot {25}^0+\tan {41}^0$