Question Number 54480 by 951172235v last updated on 04/Feb/19 $$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}\:\alpha\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{2}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{4}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\mathrm{tan}\:\left(\alpha+\frac{\mathrm{6}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:+\:\mathrm{tan}\:\left(\alpha+\frac{\mathrm{8}\overset{−} {\Lambda}}{\mathrm{5}}\right)\:=\:\mathrm{5tan}\:\mathrm{5}\alpha \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19…
Question Number 54472 by Tawa1 last updated on 04/Feb/19 $$\mathrm{If}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{are}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle},\:\:\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\:\mathrm{tanA}\:+\:\mathrm{tanB}\:+\:\mathrm{tanC}\:\:=\:\:\mathrm{tanA}\:\mathrm{tanB}\:\mathrm{tanC} \\ $$ Answered by math1967 last updated on 04/Feb/19 $${A}+{B}+{C}=\pi \\ $$$$\Rightarrow{A}+{B}=\pi−{C} \\…
Question Number 185523 by Rupesh123 last updated on 23/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185511 by mathlove last updated on 23/Jan/23 Answered by Frix last updated on 23/Jan/23 $$\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}=\mathrm{2}+\sqrt{\mathrm{3}}={t} \\ $$$${t}^{−\mathrm{cos}\:{x}} ={t}^{\mathrm{sin}\:{x}} \\ $$$$−\mathrm{cos}\:{x}\:=\mathrm{sin}\:{x} \\ $$$$\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=−\mathrm{1} \\…
Question Number 185486 by sonukgindia last updated on 22/Jan/23 Answered by a.lgnaoui last updated on 23/Jan/23 $$\mathrm{A}−\left(\mathrm{B}+\mathrm{C}\right)=\mathrm{0} \\ $$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{2}\left(\mathrm{B}+\mathrm{C}\right)=\mathrm{2A} \\ $$$$\mathrm{A}+\mathrm{B}−\mathrm{C}=\mathrm{A}−\left(\mathrm{B}+\mathrm{C}\right)+\mathrm{2B}=\mathrm{2B} \\ $$$$\mathrm{A}−\mathrm{B}+\mathrm{C}=\mathrm{A}−\left(\mathrm{B}+\mathrm{C}\right)+\mathrm{2C}=\mathrm{2C} \\ $$$$…
Question Number 119919 by bramlexs22 last updated on 28/Oct/20 $$\:\mathrm{sin}\:\mathrm{70}°\:\mathrm{cos}\:\mathrm{50}°\:+\:\mathrm{sin}\:\mathrm{260}°\:\mathrm{cos}\:\mathrm{280}°\:=? \\ $$ Answered by bobhans last updated on 28/Oct/20 $$\Leftrightarrow\mathrm{sin}\:\mathrm{70}°\:\mathrm{cos}\:\mathrm{50}°+\left(−\mathrm{cos}\:\mathrm{10}°\right)\left(\mathrm{sin}\:\mathrm{10}°\right)= \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{120}°+\mathrm{sin}\:\mathrm{20}°\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°= \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\…
Question Number 185426 by sonukgindia last updated on 21/Jan/23 Answered by Frix last updated on 21/Jan/23 $${u}=\mathrm{tan}\:{a}\:\wedge{v}=\mathrm{tan}\:{b}\:\Rightarrow \\ $$$$\frac{{u}+{v}}{\mathrm{1}−{uv}}=\frac{\mathrm{4}{v}}{\mathrm{3}−{v}^{\mathrm{2}} }\:\Rightarrow\:{u}=\frac{{v}}{\mathrm{3}}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3} \\ $$ Terms of Service…
Question Number 185320 by mathlove last updated on 20/Jan/23 Answered by SEKRET last updated on 20/Jan/23 $$\:\boldsymbol{\mathrm{sin}}\left(\mathrm{2}\boldsymbol{\mathrm{a}}\right)=\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{a}}\right)\:\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{a}}\right) \\ $$ Answered by SEKRET last updated on…
Question Number 54239 by 951172235v last updated on 01/Feb/19 Answered by Prithwish sen last updated on 01/Feb/19 $$=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{p}+\mathrm{q}+\mathrm{s}}\:+\frac{\mathrm{1}}{\mathrm{p}+\mathrm{q}+\mathrm{t}}}{\mathrm{1}−\:\frac{\mathrm{1}}{\left(\mathrm{p}+\mathrm{q}+\mathrm{s}\right)\left(\mathrm{p}+\mathrm{q}+\mathrm{t}\right)}}\right)\:+ \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{p}+\mathrm{r}+\mathrm{u}}\:+\frac{\mathrm{1}}{\mathrm{p}+\mathrm{r}+\mathrm{v}}}{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{p}+\mathrm{r}+\mathrm{u}\right)\left(\mathrm{p}+\mathrm{r}+\mathrm{v}\right)}}\:\right) \\ $$$$=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2p}+\mathrm{2q}+\mathrm{s}+\mathrm{t}}{\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}}…
Question Number 54156 by Tawa1 last updated on 30/Jan/19 $$\mathrm{Please}\:\mathrm{sirs},\:\:\mathrm{How}\:\mathrm{is},\:\:\:\:\:\:−\:\mathrm{cot}\left(\alpha\right)\:\:=\:\:\mathrm{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\:+\:\alpha\right)\:\:???? \\ $$ Commented by turbo msup by abdo last updated on 30/Jan/19 $${tan}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)\:=\frac{{sin}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)}{{cos}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)} \\ $$$$=\frac{{cos}\alpha}{−{sin}\alpha}\:=−\frac{\mathrm{1}}{{tan}\alpha}\:=−{cotan}\alpha\:.…