Answer:
Therefore, its false.
Step-by-step explanation:
To prove the identity of a given trigometric equation, we can prove it by showing both sides are equally.
- [tex] \rm{1 - \frac{sin^2(x)}{1}+sin(x)=\frac{-1}{csc(x)}} [/tex]
For [tex] \rm{1 - \frac{sin^2(x)}{1}+sin(x)=\frac{-1}{csc(x)}} [/tex] substitute [tex] \rm{x = 1} [/tex]
- [tex] \rm{\therefore \ 1 - \frac{sin^2(1)}{1}+sin(1)=\frac{-1}{csc(1)}} \\ \rm{ 1-sin^2(1)+sin(1) = \frac{-1}{\frac{1}{sin(1)}}} \\ \rm{ 1-sin^2(1)+sin(1) = -sin(1)} [/tex]
As it shown, we can see two sides are not equal.
Therefore, its false.
