Question Number 155466 by cortano last updated on 01/Oct/21 $$\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\ldots+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)=? \\ $$ Commented by VIDDD last updated on 01/Oct/21 $${can}\:{u}\:{show}\:{u}\:{solution} \\ $$…
Question Number 24355 by ahmetbak1r last updated on 16/Nov/17 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{4}+\sqrt{\mathrm{16}+\sqrt{\mathrm{256}…..}}}}=? \\ $$ Answered by ajfour last updated on 16/Nov/17 $${S}=\sqrt{\mathrm{2}^{\mathrm{0}} +\sqrt{\mathrm{2}^{\mathrm{2}} +\sqrt{\mathrm{2}^{\mathrm{4}} +\sqrt{\mathrm{2}^{\mathrm{8}} +\sqrt{..}}}}} \\…
Question Number 89848 by 9933039645 last updated on 19/Apr/20 $${Q}\mathrm{1}.\:{If}\:{sin}\theta=\frac{\mathrm{3}}{\mathrm{5}}\:{then}\:{find}\:{the}\:{value}\:{of}\:{tan}\theta+{cot}\theta \\ $$ Commented by mahdi last updated on 19/Apr/20 $$\pm\frac{\mathrm{25}}{\mathrm{12}} \\ $$$$\mathrm{sin}\theta=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow\mid\mathrm{cos}\theta\mid=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{tan}\theta=\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}=\pm\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\mathrm{cot}\theta=\frac{\mathrm{1}}{\mathrm{tan}\theta}=\pm\frac{\mathrm{4}}{\mathrm{3}}…
Question Number 24294 by ajfour last updated on 15/Nov/17 $${Find}\:{value}\left({s}\right)\:{of}\:\boldsymbol{{x}}\:{if} \\ $$$$\:\:\mathrm{sin}\:\left[\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0}\:. \\ $$ Answered by mrW1 last updated on 15/Nov/17 $$\:\mathrm{sin}\:\left[\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}}…
Question Number 89662 by 974342176 last updated on 18/Apr/20 $${Given}\:{that}\:\mathrm{sin}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\mathrm{sin}\:{C}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{without}\:{u} \\ $$$${using}\:{calculator}\:{solve} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:\left({A}+{C}\right) \\ $$$$\left.{b}\right)\:{cos}\left({A}−{C}\right) \\ $$ Answered by TANMAY PANACEA. last updated on…
Question Number 24098 by Sudipta Jana last updated on 12/Nov/17 $$\mathrm{sin}^{−\mathrm{1}} \frac{{ax}}{{c}}+\mathrm{sin}^{−\mathrm{1}} \frac{{bx}}{{c}}=\mathrm{sin}^{−\mathrm{1}} {x}\:\:\:\:\:\left[{When}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\right] \\ $$ Answered by 951172235v last updated on…
Question Number 89599 by 20000193 last updated on 18/Apr/20 Commented by john santu last updated on 18/Apr/20 $$\left.\mathrm{4}{a}\right)\:\mathrm{tan}\:\left({A}+{C}\right)\:=\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{C}}{\mathrm{1}−\mathrm{tan}\:{A}.\mathrm{tan}\:{C}} \\ $$$$\:{case}\:\left(\mathrm{1}\right)\:=\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:\sqrt{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}.\sqrt{\mathrm{3}}}\:=\:\infty\:\left({undefined}\:\right) \\ $$$${case}\:\left(\mathrm{2}\right)\:=\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:−\:\sqrt{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}.\sqrt{\mathrm{3}}}\:=\:\frac{−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${case}\:\left(\mathrm{3}\right)\:=\:\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{3}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}.\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\…
Question Number 24042 by Tinkutara last updated on 11/Nov/17 $${f}\left({x}\right)\:=\:\mathrm{sin}\left(\mathrm{sin}^{\mathrm{2}} {x}\right)\:+\:\mathrm{cos}\left(\mathrm{sin}^{\mathrm{2}} {x}\right)\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{is} \\ $$ Commented by mrW1 last updated on 11/Nov/17 $$\left[\mathrm{1},\sqrt{\mathrm{2}}\right] \\…
Question Number 89580 by jagoll last updated on 18/Apr/20 Answered by $@ty@m123 last updated on 18/Apr/20 $$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left\{\mathrm{180}−\left({x}+\mathrm{84}+\mathrm{42}\right)\right\}} \\ $$$$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left(\mathrm{54}−{x}\right)}\:…\left(\mathrm{1}\right) \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{84}}=\frac{{BC}}{\mathrm{sin}\:\left\{\mathrm{180}−\left(\mathrm{84}+\mathrm{42}\right)\right\}} \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{84}}=\frac{{AD}}{\mathrm{sin}\:\mathrm{54}}\:…\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right),…
Question Number 154958 by n0y0n last updated on 23/Sep/21 $$\:\: \\ $$$$\:\:\mathrm{if}\:,\:\mathrm{sinA}+\mathrm{sinB}=\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\:\mathrm{then}\:\:\:\frac{\mathrm{6cosA}+\mathrm{13cosB}}{\mathrm{cosA}+\mathrm{6cosB}}=? \\ $$$$\:\: \\ $$ Commented by prakash jain last updated on…