Check the solution to "Given cos⁡(a)+cos⁡(b)=1, prove that 1−s2−t2−3s2t2=0, where s=tan⁡(a2) and..."

hadejada7x

Answered question

2022-01-15

Given

\cos (a) + \cos (b) = 1

, prove that

1 - s^{2} - t^{2} - 3 s^{2} t^{2} = 0

, where

s = \tan (\frac{a}{2})

and

t = \tan (\frac{b}{2})

Elaine Verrett

Beginner2022-01-16Added 41 answers

Alternatively:

\cos a = 2 \cos^{2} \frac{a}{2} - 1

\cos^{2} \frac{a}{2} = \frac{1}{1 + \tan^{2} \frac{a}{2}} = \frac{1}{1 + s^{2}}

\cos a + \cos b = 2 \cos^{2} \frac{a}{2} + 2 \cos^{2} \frac{b}{2} - 2 =

\frac{2}{1 + s^{2}} + \frac{2}{1 + t^{2}} - 2 = 1

RizerMix

Expert2022-01-20Added 656 answers

\frac{1 - t^{2}}{1 + t^{2}} + \frac{1 - s^{2}}{1 + s^{2}} = 1

1 + s^{2} - t^{2} - s^{2} t^{2} + 1 + t^{2} - s^{2} - s^{2} t^{2} = 1 + s^{2} + t^{2} + s^{2} t^{2}

\Rightarrow 1 - s^{2} - t^{2} - 3 s^{2} t^{2} = 0

alenahelenash

Expert2022-01-24Added 556 answers

Hint:

\frac{1 - s^{2}}{1 + s^{2}} + \frac{1 - t^{2}}{1 + t^{2}} = 1

\Leftrightarrow \frac{1 - s^{2}}{1 + s^{2}} = 1 - \frac{1 - t^{2}}{1 + t^{2}}

\Leftrightarrow (1 - s^{2}) (1 + t^{2}) = 2 t^{2} (1 + s^{2})

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