Question Number 225241 by Spillover last updated on 18/Oct/25 Answered by som(math1967) last updated on 19/Oct/25 $$\:{tan}\theta=\frac{{nsin}\alpha{cos}\alpha.{sec}^{\mathrm{2}} \alpha}{\left(\mathrm{1}−{nsin}^{\mathrm{2}} \alpha\right){sec}^{\mathrm{2}} \alpha} \\ $$$$\Rightarrow{tan}\theta=\frac{{ntan}\alpha}{{sec}^{\mathrm{2}} \alpha−{ntan}^{\mathrm{2}} \alpha} \\…
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Question Number 225197 by mr W last updated on 18/Oct/25 Commented by mr W last updated on 18/Oct/25 $${just}\:{a}\:{test}\:{for}\:{animated}\:{GIF}\:{images} \\ $$ Commented by mr W…
Question Number 225230 by Abdulazim last updated on 18/Oct/25 $$\:\:\:{tg}^{\mathrm{4}} \mathrm{10}°+{tg}^{\mathrm{4}} \mathrm{50}°+{tg}^{\mathrm{4}} \mathrm{70}°=? \\ $$ Commented by Frix last updated on 18/Oct/25 $$\mathrm{Claim}: \\ $$$${x}^{\mathrm{3}}…
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Question Number 225146 by fantastic last updated on 17/Oct/25 Commented by fantastic last updated on 17/Oct/25 $${maybe}\:{it}\:{is}\:{an}\:{easy}\:{question} \\ $$$${but}\:{i}\:{feel}\:{very}\:{happy} \\ $$$${by}\:{solving}\:{this}\:{i}\:.{e}\:{irodov} \\ $$$${q}\:{by}\:{myself} \\ $$…
Question Number 225098 by Ark55 last updated on 17/Oct/25 Answered by Rasheed.Sindhi last updated on 17/Oct/25 $$\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6} \\ $$$$\left(\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\mathrm{10}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{36} \\ $$$$\mathrm{25}{x}^{\mathrm{2}}…
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Question Number 225072 by fantastic last updated on 17/Oct/25 Answered by mr W last updated on 17/Oct/25 Commented by mr W last updated on 17/Oct/25…