Given $\displaystyle sec^2 \theta + cosec^2 \theta = \frac{1}{sin^2\theta cos^2\theta}$
Proof $\displaystyle sec^2 \theta + cosec^2 \theta = 4cosec^2 (2\theta).$
$\displaystyle cosec^2 2\theta = \dfrac{1}{sin^2 2 \theta}$
Ok, add the 4 now.
$\displaystyle 4cosec^2 2\theta = \dfrac{4}{\sin^2 2 \theta}$
Now, this can be re-written as:
$\displaystyle 4 cosec^2 2\theta = \dfrac{4}{(\sin 2\theta)(\sin 2\theta)}$
Use the double angle identity I gave you;
$\displaystyle 4 cosec^2 2\theta = \dfrac{4}{(2\sin\theta\cos\theta)(2\sin\theta\cos\ theta)}$
Can you complete it now?
I think i can but can u just confirm for me please.
Thanks.
$\displaystyle 4 cosec^2 2\theta = \dfrac{4}{(2\sin\theta\cos\theta)(2\sin\theta\cos\ theta)}$
$\displaystyle 4 cosec^2 2\theta = \frac{4}{4sin^2\theta cos^2\theta}$
and the 4s cancel.
$\displaystyle 4 cosec^2 2\theta = \frac{1}{sin^2\theta cos^2\theta}$
given $\displaystyle sec^2 \theta + cosec^2 \theta = \frac{1}{sin^2\theta cos^2\theta}$
$\displaystyle sec^2 \theta + cosec^2 \theta = 4 cosec^2 2\theta$