Verify identity

[mrsrowton]

sin^2x cot^2 + sin^2x = 1

 

[Mrspi]

sin^2x cot^2 + sin^2x = 1

Both terms of the expression on the left side have a factor of sin[sup:2zs678f2]2[/sup:2zs678f2] x. Start by removing that common factor:

sin[sup:2zs678f2]2[/sup:2zs678f2] x (cot[sup:2zs678f2]2[/sup:2zs678f2] x + 1) = 1

Look at your list of identities...you should find one that gives you an expression you can use in place of (cot[sup:2zs678f2]2[/sup:2zs678f2] x + 1).....

See if that substitution helps you continue.....

 

[mrsrowton]

so i found the identidy, its csc^2x

But now i don't know what to do. I could change sin^2x to cos^2x-1

 

[Deleted member 4993]

so i found the identidy, its csc^2x

But now i don't know what to do. I could change sin^2x to cos^2x-1

Use

\(\displaystyle \ cosec(x) \ \ = \ \ \frac{1}{\sin(x)}\)

 

[mrsrowton]

ok, that was my second option.
so now i have:
sin^2x (1/sin)
now i need to cancel my sin, leaves me with:
sinx=1


Is this right? I mean I know this can not be my final answer becasue 1 has to = 1

 

[Deleted member 4993]

sin^2x cot^2 + sin^2x = 1

\(\displaystyle Left-Hand-Side (LHS)\ \ = \ \ \sin^2(x)\cdot cot^2(x) \ \ + \ \ \sin^2(x)\)

\(\displaystyle \ \ = \ \ \sin^2(x)\cdot [cot^2(x) \ \ + \ \ 1]\)

\(\displaystyle \ \ = \ \ \sin^2(x)\cdot cosec^2(x)\)

\(\displaystyle \ \ = \ \ \sin^2(x)\cdot \frac {1}{sin^2(x)}\)

\(\displaystyle \ \ = \ \ 1 = RHS\)