Question Number 224672 by wongb1506 last updated on 25/Sep/25 Answered by mr W last updated on 26/Sep/25 Commented by mr W last updated on 26/Sep/25…
Question Number 224606 by gregori last updated on 21/Sep/25 $$\:\:\:\:\mathrm{Y} \\ $$ Answered by som(math1967) last updated on 21/Sep/25 $$\:\frac{\mathrm{10sin}\:\alpha\mathrm{sin}\:\beta}{{cos}\alpha{cos}\beta}=\mathrm{9} \\ $$$$\Rightarrow\frac{{cos}\alpha{cos}\beta}{{sin}\alpha{sin}\beta}=\frac{\mathrm{10}}{\mathrm{9}} \\ $$$$\Rightarrow\frac{{cos}\alpha{cos}\beta+{sin}\alpha{sin}\beta}{{cos}\alpha{cos}\beta−{sin}\alpha{sin}\beta}=\frac{\mathrm{10}+\mathrm{9}}{\mathrm{10}−\mathrm{9}} \\…
Question Number 224562 by gregori last updated on 19/Sep/25 $$\:\:\: \\ $$ Commented by Frix last updated on 19/Sep/25 $${r}=\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\frac{{a}_{\mathrm{3}} }{{a}_{\mathrm{2}} }\:\Leftrightarrow\:\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}\:\Leftrightarrow\:\frac{\mathrm{cos}^{\mathrm{3}}…
Question Number 224515 by Vibhor last updated on 15/Sep/25 $$\frac{\mathrm{sin}\:\mathrm{35}}{\mathrm{tan}\:\mathrm{56}} \\ $$ Answered by fantastic last updated on 15/Sep/25 $$\frac{\mathrm{sin}\:\mathrm{35}^{\mathrm{0}} }{\mathrm{tan}\:\mathrm{56}^{\mathrm{0}} }=\mathrm{sin}\:\mathrm{35}^{\mathrm{0}} \mathrm{cot}\:\mathrm{56}^{\mathrm{0}} \\ $$…
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Question Number 223673 by gregori last updated on 02/Aug/25 Answered by som(math1967) last updated on 02/Aug/25 $$\:{tan}^{\mathrm{2}} {x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\frac{{sin}^{\mathrm{8}} {x}}{\mathrm{8}}+\frac{\mathrm{cos}\:^{\mathrm{8}} {x}}{\mathrm{27}}=\frac{\mathrm{1}}{\mathrm{125}} \\ $$$$\:\frac{{sin}^{\mathrm{4}} {x}}{\mathrm{2}}\:+\frac{\left(\mathrm{1}−{sin}^{\mathrm{2}}…
Question Number 222685 by gabthemathguy25 last updated on 05/Jul/25 Answered by mr W last updated on 05/Jul/25 $$\mathrm{tan}\:\left({A}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{7}}\right) \\ $$$$\frac{\mathrm{tan}\:{A}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{tan}\:{A}}=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{7}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{7}}} \\ $$$$\Rightarrow\mathrm{tan}\:{A}=\frac{\mathrm{4}}{\mathrm{3}}\:\checkmark…
Question Number 222638 by Mingma last updated on 03/Jul/25 Answered by gabthemathguy25 last updated on 03/Jul/25 $$\mathrm{cot}\:{A}\:+\:\mathrm{cot}\:{B}\:+\:\mathrm{cot}\:{C}\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}{S}} \\ $$$${where}\:{a},{b},{c}\:=\:{side}\:{lengths},\:{S}={area}\:{of}\:{triangle} \\ $$$${S}={r}\centerdot{s}\:\Rightarrow\:\frac{\mathrm{1}}{{S}}=\frac{\mathrm{1}}{{rs}} \\…
Question Number 222577 by Lekhraj last updated on 30/Jun/25 Answered by parthasc last updated on 30/Jun/25 Commented by Lekhraj last updated on 30/Jun/25 $$\mathrm{Thank}\:\mathrm{you}\:. \\…
Question Number 222279 by fantastic last updated on 21/Jun/25 $$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta+\mathrm{2}{ab}\mathrm{cos}\:\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=?? \\ $$ Answered by mr W last updated on 21/Jun/25…