I am in gr.11 just starting trig identities. I have finished all of my hw except for one question. Which looks to me like it should be simple. I just can't seem to do it. haha i'm probably just being blonde but i'll keep trying! Thanks!
I am in gr.11 just starting trig identities. I have finished all of my hw except for one question. Which looks to me like it should be simple. I just can't seem to do it. haha i'm probably just being blonde but i'll keep trying! Thanks!
$\displaystyle \text{LHS }= \csc^2(x) + \sec^2(x) = \frac{1}{\sin^2(x)} + \frac{1}{\cos^2(x)} $
$\displaystyle = \frac{\cos^2(x)}{\sin^2(x)\cos^2(x)} + \frac{\sin^2(x)}{\sin^2(x)\cos^2(x)} $
$\displaystyle = \frac{\cos^2(x) + \sin^2(x)}{\sin^2(x)\cos^2(x)} $
$\displaystyle = \frac{1}{\sin^2(x)\cos^2(x)}$
$\displaystyle = \csc^2(x)\sec^2(x) = \text{RHS} $
$\displaystyle \csc^2{x} + \sec^2{x} = \frac{1}{\sin^2{x}} + \frac{1}{\cos^2{x}}$
$\displaystyle = \frac{\cos^2{x}}{\sin^2{x}\cos^2{x}} + \frac{\sin^2{x}}{\sin^2{x}\cos^2{x}}$
$\displaystyle = \frac{\cos^2{x} + \sin^2{x}}{\sin^2{x}\cos^2{x}}$
$\displaystyle = \frac{1}{\sin^2{x}\cos^2{x}}$
$\displaystyle = \csc^2{x}\sec^2{x}$.