Proving this trigonometry proves difficult

Question

At first the question appear cheap to me, so I hesitated in solving it but when I finally tried to solve it. I did not arrive at the supposed answer

Given that $\tan A/2 = \csc A - \sin A $ Show that $\tan (A/2)^2 = -2 ± \sqrt{5}$

Using the $t$-formula to solve it, I got hooked at $3{t^4} - 4{t^2} +1 = 0$ instead of $t^4 + 4t^2 - 1 = 0$

I need help on this. How do I unravel the mystery behind this?

The solution in the handout here begin with changing to t formula Where $\tan (A/2) = t$ and $\sin A = 2t/(1 + t^2)$. We have $t = (1 + t^2)/2t - 2t/(1+t^2)$ $t = ((1 + t^2)^2 - 4t^2)/2t(1 + t^2)$

From there the handout jumped to $t^4 + 4t^2 - 1 = 0$

My problem is how to arrive at this equation above?

Welcome to Math SE! Please remember to use dollar signs when formatting ($x^2$ becomes $x^2$). — Toby Mak, Jun 16 '19 at 23:42
Show details of derivation. It is obviously a sign error someplace. — herb steinberg, Jun 16 '19 at 23:44
Thank you I'll take that to heart and thanks for the editing. — Michael Umande, Jun 16 '19 at 23:44

herb steinberg · Accepted Answer · 2019-06-17T02:01:19.977

1

$t=\frac{1+t^2}{2t}-\frac{2t}{1+t^2}=\frac{(1+t^2)^2-4t^2}{2t(1+t^2)}=\frac{(1-t^2)^2}{2t(1+t^2)}$

Clearing fraction to get $2t^2(1+t^2)=(1-t^2)^2$

or $2t^2+2t^4=1-2t^2+t^4$ or $t^4+4t^2-1=0$

You made a mistake for $sinA$ in the equation.

edited Jun 17 '19 at 02:01

answered Jun 17 '19 at 01:54

herb steinberg

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Proving this trigonometry proves difficult

1 Answers1