Question Number 164488 by mathlove last updated on 18/Jan/22 $$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{30}}−\left(\mathrm{cot}\:\mathrm{30}\right)^{\mathrm{4}} −\left(\mathrm{cot}\:\mathrm{30}\right)^{\mathrm{2}} =…. \\ $$$$\left.{a}\left.\right)\mathrm{sin}\:^{\mathrm{2}} \mathrm{30}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\right)\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{30}} \\ $$$$\left.{c}\left.\right)\mathrm{2sin}\:^{\mathrm{2}} \mathrm{30}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}\right)\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{30}} \\ $$ Commented by cortano1 last…
Question Number 164483 by mathlove last updated on 18/Jan/22 $$\frac{{cos}\mathrm{25}−{sin}\mathrm{65}}{{sin}\mathrm{20}+{sin}\mathrm{10}}=? \\ $$ Commented by som(math1967) last updated on 18/Jan/22 $$\mathrm{0} \\ $$ Answered by Rasheed.Sindhi…
Question Number 164426 by cortano1 last updated on 17/Jan/22 $$\:\:{In}\:\Delta{ABC}\:{if}\:\begin{cases}{\mathrm{cot}\:{A}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{cot}\:{B}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{18}}}\end{cases} \\ $$$$\:{then}\:\mathrm{tan}\:\:{C}\:=\:? \\ $$ Commented by bobhans last updated on 17/Jan/22 $$\:\mathrm{let}\:\mathrm{tan}\:\mathrm{C}=\mathrm{x}>\mathrm{0}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{18}}{\mathrm{x}}+\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:.\frac{\mathrm{18}}{\mathrm{x}}.\:\mathrm{x} \\…
Question Number 164391 by bobhans last updated on 16/Jan/22 $$\:\:\mathrm{If}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{cases};\:\mathrm{0}<\mathrm{A}<\frac{\pi}{\mathrm{4}}\:;\:\mathrm{0}<\mathrm{B}<\frac{\pi}{\mathrm{4}} \\ $$$$\:\mathrm{Find}\:\mathrm{tan}\:\mathrm{2A}. \\ $$ Answered by cortano1 last updated on 16/Jan/22 $$\:\:\left.\begin{matrix}{\mathrm{sin}\:\left({A}−{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{matrix}\right\}\:\Rightarrow\begin{cases}{\mathrm{0}<{A}+{B}<\frac{\pi}{\mathrm{2}}}\\{−\frac{\pi}{\mathrm{4}}<{A}−{B}<\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{tan}\:\left({A}−{B}\right)=\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{tan}\:\left({A}+{B}\right)=\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\…
Question Number 33116 by abdo imad last updated on 10/Apr/18 $${solve}\:{at}\:\left[\mathrm{0},\pi\right]\:\:{cos}\alpha\:+{cos}\left(\mathrm{2}\alpha\right)\:+{cos}\left(\mathrm{3}\alpha\right)=\mathrm{0} \\ $$ Answered by MJS last updated on 10/Apr/18 $$\mathrm{cos}\:\mathrm{2}\alpha=\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{3}\alpha=\mathrm{4cos}^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha…
Question Number 32968 by diyatrivedi last updated on 08/Apr/18 Answered by MJS last updated on 08/Apr/18 $$\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta}=\mathrm{cos}\:\alpha×\mathrm{cos}\:\theta+\mathrm{sin}\:\alpha×\mathrm{sin}\:\theta \\ $$$$\mathrm{sin}\:\alpha\:=\mathrm{cos}\:\alpha×\mathrm{cos}\:\theta×\mathrm{sin}\:\theta+\mathrm{sin}\alpha×\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\mathrm{sin}\:\alpha=\mathrm{cos}\:\alpha×\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}+\mathrm{sin}\:\alpha×\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{2}}=\mathrm{cos}\:\alpha×\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}−\mathrm{sin}\:\alpha×\frac{\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{2}} \\…
Question Number 163936 by cortano1 last updated on 12/Jan/22 $$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$ Answered by mahdipoor last updated on 12/Jan/22 $${get}\:{cos}\beta=\frac{{x}^{\mathrm{2}}…
Question Number 163926 by cortano1 last updated on 12/Jan/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\left.\begin{matrix}{\mathrm{sin}\:\theta=\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}\\{\mathrm{csc}\:^{\mathrm{2}} \theta−\mathrm{tan}\:^{\mathrm{2}} \theta\:=?\:}\end{matrix}\right\} \\ $$ Answered by MJS_new last updated on 12/Jan/22 $${s}^{\mathrm{2}} =\mathrm{1}−{s} \\…
Question Number 32841 by alimudinade@gmail.com last updated on 03/Apr/18 $${A}+{B}+{C}=\mathrm{180}^{\mathrm{0}} \\ $$$$\mathrm{3}{sin}\:{A}\:+\mathrm{4}{cosB}\:=\mathrm{6} \\ $$$$\mathrm{3}{cosA}\:+\:\mathrm{4}{sin}\:{B}\:=\sqrt{\mathrm{13}} \\ $$$${sin}\:{C}=…. \\ $$ Answered by hknkrc46 last updated on 04/Apr/18…
Question Number 163798 by kdaramaths last updated on 10/Jan/22 $$ \\ $$$$\:\:\:\:{by}\:{a}\:{simple}\:{method}\:{calculate}\:\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{k}=\mathrm{0}} ^{{n}} {COS}\left({kx}\right) \\ $$$$ \\ $$$$ \\ $$ Answered…