Question Number 17068 by chux last updated on 30/Jun/17 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$ \\ $$$$\mathrm{4tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=\pi/\mathrm{4} \\ $$ Answered by ajfour last updated on 30/Jun/17…
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Question Number 16855 by Tinkutara last updated on 27/Jun/17 $$\mathrm{If}\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{cos}\:\phi\:=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{then}\:\theta\:+\:\phi \\ $$$$\mathrm{belongs}\:\mathrm{to},\:\mathrm{where}\:\mathrm{0}\:<\:\theta,\:\phi\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\left(\frac{\pi}{\mathrm{3}},\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(\frac{\pi}{\mathrm{2}},\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$ Answered by ajfour last updated on 27/Jun/17…
Question Number 16845 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Jun/17 Commented by prakash jain last updated on 28/Jun/17 $$\left.\mathrm{1}\right) \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\mathrm{cos}\:{C} \\ $$$${C}=\mathrm{90}°,{A}=\mathrm{45}°,{B}=\mathrm{45}° \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\centerdot\mathrm{cos}\:{C}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\…
Question Number 16834 by sushmitak last updated on 26/Jun/17 $$\mathrm{If}\:\alpha<\beta<\gamma<\mathrm{2}\pi\:\mathrm{and} \\ $$$$\mathrm{cos}\:\left({x}+\alpha\right)+\mathrm{cos}\:\left({x}+\beta\right)+\mathrm{cos}\:\left({x}+\gamma\right)=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{all}\:{x}\in\mathbb{R},\:\mathrm{then}\:\mathrm{is} \\ $$$$\gamma−\alpha=\frac{\mathrm{2}\pi}{\mathrm{3}}? \\ $$ Commented by ajfour last updated on 27/Jun/17…
Question Number 16807 by sushmitak last updated on 26/Jun/17 $$\left({a}+\mathrm{2}\right)\mathrm{sin}\:\alpha+\left(\mathrm{2}{a}−\mathrm{1}\right)\mathrm{cos}\:\alpha=\left(\mathrm{2}{a}+\mathrm{1}\right) \\ $$$$\mathrm{then}\:\mathrm{tan}\:\alpha=? \\ $$ Commented by prakash jain last updated on 26/Jun/17 $$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}={u},\mathrm{sin}\:\alpha=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} },\mathrm{cos}\:\alpha=\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}}…
Question Number 16685 by rish@bh last updated on 25/Jun/17 $$\mathrm{If}\:\mathrm{cos}\:\left(\theta−\alpha\right),\mathrm{cos}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\left(\theta+\alpha\right) \\ $$$$\:\mathrm{are}\:\mathrm{in}\:\mathrm{HP},\:\mathrm{then} \\ $$$$\mathrm{cos}\:\theta\mathrm{sec}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left.\mathrm{a}\right)\pm\sqrt{\mathrm{2}} \\ $$$$\left.{b}\right)\:\pm\sqrt{\mathrm{3}} \\ $$$$\left.{c}\right)\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left.{d}\right)\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$ Answered…
Question Number 16681 by rish@bh last updated on 25/Jun/17 $$\mathrm{If} \\ $$$$\mathrm{tan}\:{x}+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{3}}+{x}\right)+\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+{x}\right)=\mathrm{3} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:{x}=\mathrm{1} \\ $$$$\left.{b}\right)\:\mathrm{tan}\:\mathrm{2}{x}=\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{tan}\:\mathrm{3}{x}=\mathrm{1} \\ $$$$\left.{d}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{above} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com…
Question Number 16675 by Tinkutara last updated on 25/Jun/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{graph}\:\mathrm{for}\:\mathrm{sin}\:{x}\:=\:\frac{{x}}{\mathrm{10}}\:\mathrm{for}\:{x}\:\in\:\left[−\pi,\:\pi\right] \\ $$$$\mathrm{is} \\ $$ Answered by ajfour last updated on 25/Jun/17 Commented by…
Question Number 82134 by Raxreedoroid last updated on 18/Feb/20 Commented by Raxreedoroid last updated on 18/Feb/20 $$\mathrm{if}\:{s}\:\mathrm{is}\:\mathrm{an}\:\mathrm{arc}\:\mathrm{for}\:\mathrm{Circle}\:\mathrm{C} \\ $$$$\mathrm{and}\:\mathrm{r}\:\mathrm{and}\:\mathrm{Radius}\:\mathrm{are}\:\mathrm{Radius}\:\mathrm{for}\:\mathrm{C} \\ $$$$\mathrm{then}… \\ $$$${s}=\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$$${d}=\frac{\mathrm{7}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)}{\mathrm{2}}…