Question

${\sin }^n{1890}^{\circ }+cose{c}^n{1890}^{\circ }=$
[$n\in N$]

• A. $n$
• B. $2n$
• C. 1
• D. 2

Answer 1

The correct option is D 2

When we see the angle more than ${360}^{\circ }$, we write those angles as $n\times ({360}^{\circ }+\theta )$, where n $ϵ$ N

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

$\Rightarrow \sin {1890}^{\circ }=\sin (360\times 5+{90}^{\circ })=\sin {90}^{\circ }=1$

and $cosecc{1890}^{\circ }=cosec(360\times 5+{90}^{\circ })=cosec{90}^{\circ }=1$

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

Thus, ${\sin }^n{1890}^{\circ }+cose{c}^n{1890}^{\circ }={\sin }^n{90}^{\circ }+cose{c}^n{90}^{\circ }$
$=1+1=2$