Author Topic: Further Trig (Read 593 times) Tweet Share

david.ko3 · « **on:** March 31, 2019, 02:20:13 pm »

Hello! How would I do this question?
Given
$x=sec\theta+tan\theta$
show that:
$\frac{x^2-1}{x^2+1}=sin\theta$
Thanks so much!!!

fun_jirachi · « **Reply #1 on:** March 31, 2019, 02:48:06 pm »

$\begin{align*}LHS &= \frac{(\sec\theta+\tan\theta)^2-1}{(\sec\theta+\tan\theta)^2+1} \\&= \frac{(\sec\theta)^2+(\tan\theta)^2+2\sec\theta\tan\theta-1}{(\sec\theta)^2+(\tan\theta)^2+2\sec\theta\tan\theta+1} \\&= \frac{2(\tan\theta)^2+2\sec\theta\tan\theta}{2(\sec\theta)^2+2\sec\theta\tan\theta} \\&= \frac{\tan\theta(\tan\theta+\sec\theta)}{\sec\theta(\tan\theta+\sec\theta)} \\&= \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}} \\&= \sin\theta \\&= RHS\end{align*}$

It's just a crap ton of algebra. Usually, when there's a show that, first thing to try would be just algebra bashing since it normally yields you the answer straight away. (Try this next time for a similar question!) Hope this helps

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Author Topic: Further Trig (Read 593 times) Tweet Share

david.ko3

Further Trig

fun_jirachi

Re: Further Trig

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