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Question Number 195578 by sonukgindia last updated on 05/Aug/23

Answered by Frix last updated on 05/Aug/23

$$\sin \alpha +\cos \alpha =\frac{\sqrt{3}}{3}$$$$t=\tan \alpha$$$$\frac{t+1}{\sqrt{t^2+1}}=\frac{\sqrt{3}}{3}$$$$t=-\frac{3-\sqrt{5}}{2}\Rightarrow$$$$t^2=\frac{7-3\sqrt{5}}{2}\wedge \frac{1}{t^2}=\frac{7+3\sqrt{5}}{2}\wedge t^2+\frac{1}{t^2}=7$$

Answered by mr W last updated on 05/Aug/23

$$\sin x+\cos x=\frac{1}{\sqrt{3}}$$$$1+2\sin x\cos x={\left(\frac{1}{\sqrt{3}}\right)}^2=\frac{1}{3}$$$$\Rightarrow \sin x\cos x=-\frac{1}{3}$$$${\tan }^{2}x+{\cot }^{2}x$$$$={(\tan x+\frac{1}{\tan x})}^2-2$$$$={\left(\frac{1}{\sin x\cos x}\right)}^2-2$$$$={\left(\frac{1}{-\frac{1}{3}}\right)}^2-2$$$$=7$$

Commented by Frix last updated on 05/Aug/23

$$\tan x+\frac{1}{\tan x}=\frac{1}{\sin x\cos x}$$$$\frac{1}{{\left(\sin x\cos x\right)}^2}-2=\frac{1}{{(-\frac{1}{3})}^2}-2=9-2=7$$

Commented by mr W last updated on 05/Aug/23

$$yes,thanks!$$

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