Question

In $△ABC$ , $\angle B={90}^{\circ }$.

If $cotA=\frac{1}{\sqrt{3}}$ then,

find $\frac{cosA}{sinA}+\frac{sinC}{cosC}$

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{2}{\sqrt{3}}$

Given:
$\angle$ $B$ = ${90}^{\circ }$ and cot $A$ = $\frac{1}{\sqrt{3}}$

$\because$ cot ${60}^{\circ }$ = $\frac{1}{\sqrt{3}}$
$\therefore$ $A$ = ${60}^{\circ }$

$\because$ Sum of all angles of a triangle = ${180}^{\circ }$
$\therefore$ $\angle$ $A$ + $\angle$ $B$ + $\angle$ $C$ = ${180}^{\circ }$
$\Rightarrow$ ${60}^{\circ }$ + ${90}^{\circ }$ + $\angle$ $C$ = ${180}^{\circ }$
$\Rightarrow$ $\angle$ $C$ = ${180}^{\circ }$ - ${60}^{\circ }$ - ${90}^{\circ }$
$\therefore$ $\angle$ $C$ = ${30}^{\circ }$

Now,

$\frac{cosA}{sinA}$ = cot $A$ = $\frac{1}{\sqrt{3}}$

and

$\frac{sinC}{cosC}$ = tan $C$ = tan ${30}^{\circ }$ = $\frac{1}{\sqrt{3}}$

$\therefore$ $\frac{cosA}{sinA}$ + $\frac{sinC}{cosC}$ = cot A + tan C

$\therefore$$\frac{cosA}{sinA}+\frac{sinC}{cosC}=\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$