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May 11, 2021, 07:45:16 am

### AuthorTopic: Further Trig  (Read 327 times) Tweet Share

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#### david.ko3

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##### Further Trig
« on: March 31, 2019, 02:20:13 pm »
0
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

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##### Re: Further Trig
« Reply #1 on: March 31, 2019, 02:48:06 pm »
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\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
« Last Edit: March 31, 2019, 04:56:58 pm by fun_jirachi »
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