Question

If $\tan {32}^0.\cot ({90}^0-\theta )=1$ find $\theta$.

• A. ${58}^{\circ }$
• B. ${122}^{\circ }$
• C. ${32}^{\circ }$
• D. ${158}^{\circ }$

Answer 1

The correct option is A ${58}^{\circ }$

Given, $\tan {32}^0.\cot ({90}^0-\theta )=1$
$\Rightarrow \tan {32}^0=\frac{1}{\cot ({90}^0-\theta )}$
$\Rightarrow \tan {32}^0=\tan ({90}^0-\theta )$
$\Rightarrow ({90}^0-\theta )={32}^0$
$\Rightarrow \theta ={90}^0-{32}^0$

$={58}^0$
Option A is correct.