Question Number 165649 by cortano1 last updated on 06/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 165637 by daus last updated on 05/Feb/22 $$\mathrm{37}{tanx}\:=\mathrm{11}{tan}\mathrm{3}{x} \\ $$$${solve}\:{it} \\ $$ Answered by mr W last updated on 06/Feb/22 $$\mathrm{37}\:\mathrm{tan}\:{x}=\mathrm{11}×\frac{\mathrm{3}\:\mathrm{tan}\:{x}−\mathrm{tan}^{\mathrm{3}} \:{x}}{\mathrm{1}−\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:{x}}…
Question Number 165558 by som(math1967) last updated on 03/Feb/22 $${Prove}\:{that} \\ $$$$\:\:\boldsymbol{{sec}}\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{7}}\:+\boldsymbol{{sec}}\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{7}}+\boldsymbol{{sec}}\frac{\mathrm{8}\boldsymbol{\pi}}{\mathrm{7}}\:=−\mathrm{4} \\ $$ Answered by Rohit143Jo last updated on 04/Feb/22 $${Ans}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{Let},\:{x}=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:\:\:\Leftrightarrow\:\:\mathrm{7}{x}=\mathrm{2}\pi \\…
Question Number 34420 by math khazana by abdo last updated on 06/May/18 $${let}\:{A}\:\:=\:{arctanx}\:−{arctany} \\ $$$${give}\:{another}\:{form}\:{of}\:{A}\:{if}\:\:{xy}\neq−\mathrm{1}\:. \\ $$ Commented by math khazana by abdo last updated…
Question Number 165428 by mathlove last updated on 01/Feb/22 Answered by som(math1967) last updated on 01/Feb/22 $$\:\frac{{sin}\mathrm{6}{sin}\mathrm{42}{sin}\mathrm{66}{sin}\mathrm{78}}{{cos}\mathrm{6}{cos}\mathrm{42}{cos}\mathrm{66}{cos}\mathrm{78}}\:\:\bigstar \\ $$$$=\frac{\left(\mathrm{2}{sin}\mathrm{6}{sin}\mathrm{66}\right)\left(\mathrm{2}{sin}\mathrm{42}{sin}\mathrm{78}\right)}{\left(\mathrm{2}{cos}\mathrm{6}{cos}\mathrm{66}\right)\left(\mathrm{2}{cos}\mathrm{42}{cos}\mathrm{78}\right)} \\ $$$$=\frac{\left\{{cos}\left(\mathrm{66}−\mathrm{6}\right)−{cos}\left(\mathrm{66}+\mathrm{6}\right)\right\}\left\{{cos}\mathrm{36}−{cos}\mathrm{120}\right\}}{\left({cos}\mathrm{72}+{cos}\mathrm{60}\right)\left({cos}\mathrm{120}+{cos}\mathrm{36}\right)} \\ $$$$=\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\:\:\bigstar\bigstar \\ $$$$=\frac{\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}=\frac{\mathrm{9}−\mathrm{5}}{\mathrm{5}−\mathrm{1}}=\mathrm{1}…
Question Number 165382 by mathlove last updated on 31/Jan/22 Commented by mathlove last updated on 01/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 165370 by daus last updated on 31/Jan/22 Commented by daus last updated on 31/Jan/22 $${solve}\:{x} \\ $$ Answered by Ar Brandon last updated…
Question Number 165314 by mathlove last updated on 29/Jan/22 $$\mathrm{sin}\:^{\mathrm{2}} \mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}+…..+{sin}^{\mathrm{2}} \mathrm{90}=? \\ $$ Commented by cortano1 last updated on 29/Jan/22 $$\mathrm{45}.\mathrm{5}\: \\…
Question Number 99742 by Ar Brandon last updated on 23/Jun/20 $$\mathcal{G}\mathrm{iven}\:\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}°\:\:\mathrm{prove}\:\:\mathrm{that} \\ $$$$\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}+\mathrm{tan}\frac{\mathrm{C}}{\mathrm{2}}\mathrm{tan}\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{1} \\ $$ Answered by smridha last updated on 23/Jun/20 $$\boldsymbol{{tan}}\left(\frac{\boldsymbol{{A}}}{\mathrm{2}}+\frac{\boldsymbol{{B}}}{\mathrm{2}}+\frac{\boldsymbol{{C}}}{\mathrm{2}}\right)=\boldsymbol{{tan}}\frac{\boldsymbol{\pi}}{\mathrm{2}}=\infty\: \\ $$$$\Rightarrow\frac{\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}−\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}}{\mathrm{1}−\left[\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{B}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}+\boldsymbol{{tan}}\frac{\boldsymbol{{C}}}{\mathrm{2}}.\boldsymbol{{tan}}\frac{\boldsymbol{{A}}}{\mathrm{2}}\right]}=\infty…
Question Number 165270 by cortano1 last updated on 28/Jan/22 Answered by TheSupreme last updated on 28/Jan/22 $${B}+{C}=\mathrm{120} \\ $$$$\frac{{sin}\left({C}\right)}{{sin}\left({B}\right)}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${sin}\left({C}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){sin}\left(\mathrm{120}−{C}\right) \\ $$$${sin}\left({C}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\frac{{cos}\left({C}\right)}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\left({C}\right)\right) \\ $$$${sin}\left({C}\right)\left(\mathrm{1}+\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\frac{{cos}\left({C}\right)}{\mathrm{2}}…