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…
Question Number 23851 by tawa tawa last updated on 08/Nov/17 $$\mathrm{Solve}\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{83}\right)\:−\:\mathrm{cos}\left(\mathrm{17}\right)\:\:\:\:\:\:\mathrm{in}\:\mathrm{surd}\:\mathrm{form} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89351 by john santu last updated on 17/Apr/20 $${prove}\:\mathrm{tan}\:\mathrm{3}^{{o}} ×\mathrm{tan}\:\mathrm{39}^{{o}} ×\mathrm{tan}\:\mathrm{89}^{{o}} \:=\:\mathrm{tan}\:\mathrm{15}^{{o}} \\ $$ Commented by MJS last updated on 17/Apr/20 $$\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\…
Question Number 89327 by Ar Brandon last updated on 16/Apr/20 Commented by MJS last updated on 16/Apr/20 $$\mathrm{look}\:\mathrm{at}\:\mathrm{qu}.\:\mathrm{89290} \\ $$ Commented by Ar Brandon last…
Question Number 23721 by Tinkutara last updated on 06/Nov/17 $$\mathrm{If}\:\mathrm{symbols}\:\mathrm{have}\:\mathrm{their}\:\mathrm{usual}\:\mathrm{meaning} \\ $$$$\mathrm{then}\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}_{\mathrm{2}} ^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{r}_{\mathrm{3}} ^{\mathrm{2}} }\:= \\ $$$$\left(\mathrm{1}\right)\:\frac{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} }{{s}^{\mathrm{2}} }…
Question Number 23718 by vick007 last updated on 05/Nov/17 $$\mathrm{5}\left(\mathrm{tan}^{\mathrm{2}} {x}−\mathrm{cos}^{\mathrm{2}} {x}\right)=\mathrm{2}{cosx}+{a}\:{then}\:{find} \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:{cos}\mathrm{4}{x}.??? \\ $$ Commented by vick007 last updated on 04/Nov/17 $${please}\:{ans}\:{me}\:{soon}… \\…
Question Number 154748 by imjagoll last updated on 21/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 21/Sep/21 $$\mathrm{sin}\left(\mathrm{3log}_{\left(\mathrm{2sin}{x}\right)} \sqrt[{\mathrm{3}}]{\pi}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{log}_{\left(\mathrm{2sin}{x}\right)} \pi=\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{log}_{\pi} \left(\mathrm{2sin}{x}\right)=\frac{\mathrm{6}}{\pi}\:\Rightarrow\mathrm{sin}{x}=\frac{\mathrm{1}}{\mathrm{2}}\pi^{\frac{\mathrm{6}}{\pi}} \\ $$$$\Rightarrow{x}=\mathrm{arcsin}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi^{\frac{\mathrm{6}}{\pi}} \right)…