Prove that $\displaystyle sec^2 \theta + cosec^2 \theta \equiv 4cosec^2 2\theta $

Really not sure about using the compound angle formula. Thanks for any help

Printable View

- Sep 9th 2009, 12:22 PMgreghuntertrigonometry compund angle formula
Prove that $\displaystyle sec^2 \theta + cosec^2 \theta \equiv 4cosec^2 2\theta $

Really not sure about using the compound angle formula. Thanks for any help - Sep 9th 2009, 01:00 PMred_dog
$\displaystyle \sec^2\theta+\csc^2\theta=\frac{1}{\cos^2\theta}+\ frac{1}{\sin^2\theta}=\frac{\sin^2\theta+\cos^2\th eta}{\sin^2\theta\cos^2\theta}=$

$\displaystyle =\frac{1}{\sin^2\theta\cos^2\theta}=\frac{4}{4\sin ^2\theta\cos^2\theta}=\frac{4}{\sin^22\theta}=4\cs c^22\theta$ - Sep 10th 2009, 10:55 AMgreghunterFinding the values
Thanks, I managed to work out what to do in the end. Can anyone tell me what you would do to find $\displaystyle \theta$ ?

- Sep 10th 2009, 06:56 PMpacman
aaahh, you post this as a trigonometric equation? i thought it was an identity as as solved by red_dog, YOUR post is an identity . . . .