Question

If : a^2 sec^2 - b^2 tan^2 = c^2
Prove that : sin ^2 = c^2 - a^2 / c^2 - b^2

Is there any trigonometry expert out there?? plz help

Answer 1

There's an identity, tan^2 x + 1 = sec^2 x, that comes from dividing both sides of the better known sin^2 x + cos^2 x = 1 /cos^2 x.  

So if a^2 sec^2 x - b^2 tan^2 x = c^2, you deduce that a^2 (tan^2 x + 1) - b^2 tan^2 x = c^2, or that (a^2 - b^2) tan^2 x + a^2 = c^2, or that (a^2 - b^2) tan^2 x = c^2 - a^2, or that tan^2 x = (c^2 - a^2)/(a^2 - b^2).  

You could also use the original identity tan^2 x + 1 = sec^2 x to write tan^2 x = sec^2 x - 1 in terms of sec^2 x. Plugging _this_ into a^2 sec^2 x - b^2 tan^2 x = c^2, you get a^2 sec^2 x - b^2 (sec^2 x - 1) = c^2 or that (a^2 - b^2) sec^2 x + b^2 = c^2 or that (a^2 - b^2) sec^2 x = c^2 - b^2 or that sec^2 x = (c^2 - b^2)/(a^2 - b^2).  

Since tan^2 x = (sin^2 x)/(cos^2 x) = sin^2 x sec^2 x we see that sin^2 x = (tan^2 x)/(sec^2 x). Using the two formulas from the previous paragraphs we see that the a^2 - b^2 terms cancel and thus  

sin^2 x = (c^2 - a^2)/(c^2 - b^2).