Question
If $\tan \theta=\frac{x^{2}+1}{x^{2}-1}$, then(a) $\sin \theta=\frac{x^{2}+1}{\sqrt{2\left(x^{4}+1\right)}}$(b) $\sin \theta=\frac{x^{2}-1}{\sqrt{2\left(x^{4}+1\right)}}$(c) $\cos \theta=\frac{x^{2}+1}{\sqrt{2\left(x^{4}+1\right)}}$(d) $\cos \theta=\frac{x^{2}-1}{\sqrt{2\left(x^{4}+1\right)}}$.
Step 1
Let's denote the perpendicular as $a$, the base as $b$, and the hypotenuse as $c$. So, $a=x^{2}+1$ and $b=x^{2}-1$. Show more…
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