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# Answer on Trigonometry Question for Abhyuday singh

Question #2709
Solve for x: tan2x+cot2x=2
&lt;img src=&quot;/cgi-bin/mimetex.cgi?%5Ctan%5E2x+%5Ccot%5E2x%20=%202%20%5C%5C%20%5Ctan%5E2x%20+%20%5Cfrac%7B1%7D%7B%5Ctan%5E2x%7D%20=%202%20%5C%5C%20%5Cfrac%7Btan%5E4x+1%7D%7Btan%5E2x%7D%20=2%20%5C%5C%20tan%5E4x%20+1%20=%202%5Ctan%5E2x%20%5C%5C%20tan%5E4x%20-%202%5Ctan%5E2x%20+1%20=0%20%5C%5C%20%28%5Ctan%5E2x-1%29%5E2%20=%200%20%5C%5C%20%5Ctan%5E2x=1%20%5C%5C%20%5Ctan%20x%20=%201%20%5Chspace%7B3mm%7D%20%5Ctan%20x%20=%20-1%20%5C%5C%20x%20=%20%5Cpi/4%20+%20%5Cpi%20n%20%5Chspace%7B3mm%7D%20and%20%5C%20x%20=%20-%20%5Cpi/4%20+%20%5Cpi%20n%20%5C%5C%20x%20=%20%5Cpi/4%20%5Cpm%20%5Cpi%20n/2&quot; title=&quot;\tan^2x+\cot^2x = 2 \\ \tan^2x + \frac{1}{\tan^2x} = 2 \\ \frac{tan^4x+1}{tan^2x} =2 \\ tan^4x +1 = 2\tan^2x \\ tan^4x - 2\tan^2x +1 =0 \\ (\tan^2x-1)^2 = 0 \\ \tan^2x=1 \\ \tan x = 1 \hspace{3mm} \tan x = -1 \\ x = \pi/4 + \pi n \hspace{3mm} and \ x = - \pi/4 + \pi n \\ x = \pi/4 \pm \pi n/2&quot;&gt;

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