Question

Please solve:

ASAP

Urgent ​

Answer 1

We're given the equation,

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

Take $\sin\theta=x,$ then $\csc\theta=\frac{1}{x}.$ Then the equation becomes,

$\longrightarrow x+\dfrac{1}{x}+2=0$

Multiply both sides by $x,$ assuming $x\neq 0.$ 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,

$\longrightarrow y=x^{2019}+\dfrac{1}{x^{2020}}$

Now putting $x=-1,$

$\longrightarrow y=(-1)^{2019}+\dfrac{1}{(-1)^{2020}}$

$\longrightarrow y=-1+\dfrac{1}{1}$

$\longrightarrow y=-1+1$

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

Hence 0 is the answer.