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Question Number 108780 by 150505R last updated on 19/Aug/20

Answered by bobhans last updated on 19/Aug/20

$$\cot {}^{2}81°=\tan {}^{2}9$$$$\cot {}^{2}63°=\tan {}^{2}27$$$$(\ast )\frac{1}{2-\cot {}^{2}9°}+\frac{1}{2-\tan {}^{2}9°}+\frac{1}{2-\cot {}^{2}27°}+\frac{1}{2-\tan {}^{2}27°}=\mu$$$$let9°=x$$$$\mu =\frac{1}{2-\tan {}^{2}x}+\frac{1}{2-\frac{1}{\tan {}^{2}x}}+\frac{1}{2-\frac{1}{\tan {}^{2}3x}}+\frac{1}{2-\tan {}^{2}3x}$$

Commented by bobhans last updated on 19/Aug/20

$$let\tan x=q$$$$$$

Commented by 150505R last updated on 19/Aug/20

$$canisolveitfurther?$$

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