Question

Please solve:

ASAP

Urgent ​

Answer 1

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
`x,` 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.