Question

Find the value of ${\sin }^n{1890}^{\circ }+{\text{cosec}}^n{1890}^{\circ }$. Where $n\in N$

• A. $n$
• B. $2n$
• C. $1$
• D. $2$

Answer 1

The correct option is D

$2$

When we see the angle more then ${360}^{\circ }$. We will write those angles as $(n\times {360}^{\circ }+\theta )$ where $n\in N$

${1890}^{\circ }$ = ${360}^{\circ }\times 5+{90}^{\circ }$

$\sin ({360}^{\circ }\times 5+{90}^{\circ })=\sin {90}^{\circ }=1$

and $\text{cosec}({360}^{\circ }\times 5+{90}^{\circ })=\text{cosec}{90}^{\circ }=1$

So, ${\sin }^n$ $({360}^{\circ }\times 5+{90}^{\circ })=1$
$\text{cosec}{}^n$ $({360}^{\circ }\times$ 5 $+$ ${90}^{\circ })=1$

$={\sin }^n$ ${90}^{\circ }+{\text{cosec}}^n$ ${90}^{\circ }$

$=1+1$

$=2$