1. ## Trig Proof

Given $sec^2 \theta + cosec^2 \theta = \frac{1}{sin^2\theta cos^2\theta}$

Proof $sec^2 \theta + cosec^2 \theta = 4cosec^2 (2\theta).$

2. Use what you've been given and use the identity:

$\sin(2A) = 2\sin A\cos A$

Can you do it? Or still stuck?

3. still stuck tbh.

does $cosec^2 (2\theta)$ = $\frac{1}{sin^2 (2 \theta)}$
if so,
does $\frac{1}{sin^2 (2\theta)} = \frac{1}{sin^2\theta cos^2\theta} + \frac{1}{ cos^2\theta sin^2\theta}$

4. $cosec^2 2\theta = \dfrac{1}{sin^2 2 \theta}$

$4cosec^2 2\theta = \dfrac{4}{\sin^2 2 \theta}$

Now, this can be re-written as:

$4 cosec^2 2\theta = \dfrac{4}{(\sin 2\theta)(\sin 2\theta)}$

Use the double angle identity I gave you;

$4 cosec^2 2\theta = \dfrac{4}{(2\sin\theta\cos\theta)(2\sin\theta\cos\ theta)}$

Can you complete it now?

5. I think i can but can u just confirm for me please.

Thanks.

$4 cosec^2 2\theta = \dfrac{4}{(2\sin\theta\cos\theta)(2\sin\theta\cos\ theta)}$

$4 cosec^2 2\theta = \frac{4}{4sin^2\theta cos^2\theta}$

and the 4s cancel.

$4 cosec^2 2\theta = \frac{1}{sin^2\theta cos^2\theta}$

given $sec^2 \theta + cosec^2 \theta = \frac{1}{sin^2\theta cos^2\theta}$

$sec^2 \theta + cosec^2 \theta = 4 cosec^2 2\theta$

6. That's it, right!

7. (yes).