Please solve: ASAP Urgent

Question

asked Mar 15, 2023 231k views

1 Answer

Sam Plus Plus · Answer 1 · 2023-03-19T19:30:01+0000

We're given the equation,

$\longrightarrow\sin\theta+\csc\theta+2=0$

Take
$\sin\theta=x,$ then
$\csc\theta=(1)/(x).$ Then the equation becomes,

$\longrightarrow x+(1)/(x)+2=0$

Multiply both sides by
assuming
$x\\eq 0.$ Then we get,

$\longrightarrow x^2+2x+1=0$

$\longrightarrow (x+1)^2=0$

$\Longrightarrow x=-1$

We're asked to find the value of,

$\longrightarrow y=\sin^(2019)\theta+\csc^(2020)\theta$

As we've taken
$\sin\theta=x,$ this expression becomes,

$\longrightarrow y=x^(2019)+(1)/(x^(2020))$

Now putting
x=-1,

$\longrightarrow y=(-1)^(2019)+(1)/((-1)^(2020))$

$\longrightarrow y=-1+(1)/(1)$

$\longrightarrow y=-1+1$

$\longrightarrow\underline{\underline{y=0}}$

Hence 0 is the answer.