Question Number 91689 by john santu last updated on 02/May/20 $$\mathrm{sec}\:^{\mathrm{2}} \mathrm{1}^{{o}} +\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}^{{o}} +\mathrm{sec}\:^{\mathrm{2}} \mathrm{3}^{{o}} +…+\mathrm{sec}\:^{\mathrm{2}} \mathrm{89}^{{o}} \\ $$ Commented by Prithwish Sen 1…
Question Number 26135 by JI Siam last updated on 21/Dec/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{b}=\mathrm{2asin}^{\mathrm{2}} \theta\:; \\ $$$$\mathrm{when}\:\mathrm{acos}\theta−\mathrm{bsin}\theta=\mathrm{c}\:\mathrm{and}\:\theta=\mathrm{45}° \\ $$ Commented by mrW1 last updated on 21/Dec/17 $${It}\:{is}\:{not}\:{true}\:{that}\:{b}=\mathrm{2}{a}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:{if}\:{c}\neq\mathrm{0}.…
Question Number 26050 by Tinkutara last updated on 18/Dec/17 Answered by ajfour last updated on 20/Dec/17 $$\left[\mathrm{1},\infty\right) \\ $$$$\:\:{mere}\:\:{attempt}\:…… \\ $$$${b}+{c}\:\geqslant\:{a} \\ $$$${a}+{b}+{c}\:\geqslant\:\mathrm{2}{a} \\ $$$$\Rightarrow\:\frac{{s}}{{a}}\:\geqslant\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….\left({i}\right)…
Question Number 26008 by Tinkutara last updated on 17/Dec/17 Answered by ajfour last updated on 18/Dec/17 $$\left(\mathrm{3}\right) \\ $$ Answered by ajfour last updated on…
Question Number 26007 by Tinkutara last updated on 17/Dec/17 Answered by ajfour last updated on 17/Dec/17 $${r}_{\mathrm{1}} \left(\mathrm{tan}\:\frac{{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{{C}}{\mathrm{2}}\right)={a} \\ $$$${r}\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)={a} \\ $$$$\Rightarrow\:\frac{{r}_{\mathrm{1}} }{{r}}=\frac{\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}}{\mathrm{tan}\:\frac{{B}}{\mathrm{2}}+\mathrm{tan}\:\frac{{C}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{B}}{\mathrm{2}}\mathrm{tan}\:\frac{{C}}{\mathrm{2}}}…
Question Number 157065 by apriadodir last updated on 19/Oct/21 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{cos}\:\mathrm{36}°=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$ Commented by cortano last updated on 19/Oct/21 $$\:\mathrm{cos}\:\mathrm{36}°=\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{18}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\left(\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}\right)…
Question Number 91484 by MWSuSon last updated on 01/May/20 Commented by mr W last updated on 01/May/20 $${question}\:{was}\:{answered}\:{some}\:{days}\: \\ $$$${ago}.\:{please}\:{try}\:{to}\:{find}\:{the}\:{old}\:{post}. \\ $$ Commented by otchereabdullai@gmail.com…
Question Number 25842 by Tinkutara last updated on 15/Dec/17 Commented by ajfour last updated on 15/Dec/17 $$\mathrm{2sin}\:\left(\frac{{A}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{B}}{\mathrm{2}}\right)=\mathrm{2}\left(\frac{\mathrm{sin}\:{A}+\mathrm{sin}\:{B}}{\mathrm{sin}\:{C}}\right)−\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)−\mathrm{cos}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)}{\mathrm{2sin}\:\left(\frac{{C}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)}−\mathrm{2} \\ $$$$\Rightarrow\:{as}\:\:\:\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)=\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)\:,\:{so} \\ $$$$\left[\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)−\mathrm{cos}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right)\right]\mathrm{sin}\:\frac{{C}}{\mathrm{2}}…
Question Number 156887 by cortano last updated on 16/Oct/21 $$\:\mathrm{tan}\:\mathrm{2}{x}\:\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{tan}\:\mathrm{5}{x}\:=\mathrm{1} \\ $$ Answered by peter frank last updated on 18/Oct/21 $$\mathrm{tan}\:\left(\mathrm{x}+\mathrm{y}\right)=\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{tan}\:\mathrm{y}}{\mathrm{1}−\mathrm{tan}\:\mathrm{xtan}\:\mathrm{y}} \\ $$$$\mathrm{tan}\:\mathrm{5x}=\mathrm{tan}\:\left(\mathrm{2x}+\mathrm{3x}\right)=\frac{\mathrm{tan}\:\mathrm{2x}+\mathrm{tan}\:\mathrm{3x}}{\mathrm{1}−\mathrm{tan}\:\mathrm{2xtan}\:\mathrm{3y}} \\ $$$$\mathrm{tan}\:\mathrm{5x}−\mathrm{tan}\:\mathrm{5xtan}\:\mathrm{2xtan}\:\mathrm{3x}=\mathrm{tan}\:\mathrm{2x}+\mathrm{tan}\:\mathrm{3x}…
Question Number 25771 by keyurpatel last updated on 14/Dec/17 $${show}\:{that}\:\frac{\mathrm{cos}\:\mathrm{50}}{\mathrm{sin}\:\mathrm{40}}+\frac{\mathrm{sin}\:\mathrm{42}}{\mathrm{cos}\:\mathrm{48}}−\frac{\mathrm{2tan}\:\mathrm{18}}{\mathrm{cot}\:\mathrm{72}}=\mathrm{0} \\ $$ Answered by deepak123 last updated on 14/Dec/17 $$\mathrm{sin}\:\left(\mathrm{90}−\theta\right)=\mathrm{cos}\theta \\ $$$$\mathrm{cos}\:\left(\mathrm{90}−\theta\right)=\mathrm{sin}\:\theta \\ $$$$\mathrm{tan}\:\left(\mathrm{90}−\theta\right)=\mathrm{cot}\:\theta \\…