Question Number 18797 by Tinkutara last updated on 29/Jul/17 $$\mathrm{If}\:{A},\:{B},\:{C}\:\mathrm{are}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}, \\ $$$$\mathrm{then}\:\mathrm{2sin}\frac{{A}}{\mathrm{2}}\mathrm{cosec}\frac{{B}}{\mathrm{2}}\mathrm{sin}\frac{{C}}{\mathrm{2}}\:−\:\mathrm{sin}{A}\mathrm{cot}\frac{{B}}{\mathrm{2}} \\ $$$$−\:\mathrm{cos}\:{A}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Independent}\:\mathrm{of}\:{A},\:{B},\:{C} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Function}\:\mathrm{of}\:{A},\:{B},\:{C} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Function}\:\mathrm{of}\:{A},\:{B} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Function}\:\mathrm{of}\:{B},\:{C} \\ $$ Answered…
Question Number 18742 by Tinkutara last updated on 31/Jul/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cot16}°\mathrm{cot44}°\:+\:\mathrm{cot44}°\mathrm{cot76}° \\ $$$$−\:\mathrm{cot76}°\mathrm{cot16}°\:\mathrm{is} \\ $$ Answered by Tinkutara last updated on 31/Jul/17 $$\frac{\mathrm{3}\:+\:\mathrm{cot}\:\mathrm{76}°\:\mathrm{cot}\:\mathrm{16}°}{\mathrm{cot}\:\mathrm{76}°\:+\:\mathrm{cot}\:\mathrm{16}°} \\ $$$$=\:\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{76}°\:\mathrm{sin}\:\mathrm{16}°\:+\:\mathrm{cos}\:\mathrm{76}°\:\mathrm{cos}\:\mathrm{16}°}{\mathrm{sin}\:\left(\mathrm{76}°\:+\:\mathrm{16}°\right)} \\…
Question Number 18723 by mondodotto@gmail.com last updated on 28/Jul/17 Answered by Tinkutara last updated on 29/Jul/17 $$\mathrm{2sin2}\theta\:+\:\mathrm{15cos2}\theta\:=\:\mathrm{10} \\ $$$$\frac{\mathrm{2sin2}\theta}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{15cos2}\theta}{\:\sqrt{\mathrm{229}}}\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}} \\ $$$$\mathrm{cos}\left(\mathrm{2}\theta\:−\:\alpha\right)\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{229}}}\:,\:\mathrm{where}\:\alpha\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right) \\ $$$$\theta\:=\:{n}\pi\:\pm\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{229}}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{15}}{\:\sqrt{\mathrm{229}}}\right)…
Question Number 18721 by Tinkutara last updated on 28/Jul/17 $$\mathrm{Let}\:{ABC}\:\mathrm{and}\:{ABC}'\:\mathrm{be}\:\mathrm{two}\:\mathrm{non}- \\ $$$$\mathrm{congruent}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides}\:{AB}\:=\:\mathrm{4}, \\ $$$${AC}\:=\:{AC}'\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{angle}\:{B}\:=\:\mathrm{30}°. \\ $$$$\mathrm{The}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{is} \\ $$ Answered by ajfour last updated…
Question Number 18719 by Tinkutara last updated on 28/Jul/17 $$\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{1}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{2}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{3}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{4}} \\ $$$$+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{5}} \:=\:\mathrm{5},\:\mathrm{then}\:\mathrm{sin}\:{x}_{\mathrm{1}} \:+\:\mathrm{2sin}\:{x}_{\mathrm{2}} \:+ \\ $$$$\mathrm{3sin}\:{x}_{\mathrm{3}}…
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Question Number 18695 by Arnab Maiti last updated on 27/Jul/17 $$\mathrm{if}\:\:\frac{\mathrm{n}\:\mathrm{tan}\theta}{\mathrm{cos}^{\mathrm{2}} \left(\alpha−\theta\right)}=\frac{\mathrm{m}\:\mathrm{tan}\left(\alpha−\theta\right)}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{2}\theta=\alpha−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{n}−\mathrm{m}}{\mathrm{n}+\mathrm{m}}\mathrm{tan}\alpha\right) \\ $$ Answered by Tinkutara last updated…
Question Number 18693 by Arnab Maiti last updated on 27/Jul/17 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{tan}\frac{\alpha}{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\alpha\:\mathrm{cos}\beta}{\mathrm{cos}\alpha\:+\mathrm{sin}\beta}\right) \\ $$ Answered by 951172235v last updated on 01/Feb/19 $$\mathrm{tan}\:\frac{\Theta}{\mathrm{2}\:}\:=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\left(\frac{\overset{−}…
Question Number 84136 by mathocean1 last updated on 09/Mar/20 $${show}\:{that}: \\ $$$${tan}\mathrm{3}{x}=\frac{\mathrm{3}+{tan}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {x}}×{tanx} \\ $$ Answered by Rio Michael last updated on 09/Mar/20 $$\mathrm{tan3}{x}\:=\:\mathrm{tan}\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\frac{\mathrm{tan2}{x}\:+\:\mathrm{tan}{x}}{\mathrm{1}−\mathrm{tan2}{x}\mathrm{tan}{x}}…
Question Number 18588 by Joel577 last updated on 25/Jul/17 $$\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{a}\:+\:\mathrm{3}}{{a}\:+\:\mathrm{1}} \\ $$$$\mathrm{How}\:\mathrm{many}\:{a}\:\mathrm{that}\:\mathrm{can}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{above}? \\ $$ Answered by ajfour last updated on 25/Jul/17 $$−\mathrm{1}\leqslant\frac{\mathrm{2a}+\mathrm{3}}{\mathrm{a}+\mathrm{1}}\leqslant\mathrm{1}\:\:\:;\:\:\mathrm{a}\neq−\mathrm{1} \\…