Question Number 161903 by amin96 last updated on 23/Dec/21 $$\frac{\mathrm{1}−\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}=??? \\ $$$$ \\ $$ Commented by blackmamba last updated on 24/Dec/21 $$\:=\frac{\mathrm{1}−\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}\right)}{\mathrm{1}+\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\…
Question Number 30823 by drishtantsingh315 last updated on 26/Feb/18 Answered by math1967 last updated on 27/Feb/18 $${Cos}\left[\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{sin}\:\left(\mathrm{cot}^{−\mathrm{1}} {x}\right)\right\}\right] \\ $$$$\mathrm{cos}\:\left[\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{sin}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}\right)\right\}\right] \\…
Question Number 96282 by 675480065 last updated on 31/May/20 Answered by bemath last updated on 31/May/20 $$\mathrm{look}\:\mathrm{qn}\:\mathrm{94383} \\ $$ Answered by 1549442205 last updated on…
Question Number 96260 by bemath last updated on 31/May/20 $$\mathrm{tan}\:\left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{7}\pi}{\mathrm{9}}\right)\:=? \\ $$ Answered by john santu last updated on 31/May/20 $$\mathrm{set}\:{x}\:=\:\frac{\pi}{\mathrm{9}},\:\frac{\mathrm{4}\pi}{\mathrm{9}},\:\frac{\mathrm{7}\pi}{\mathrm{9}} \\ $$$$\mathrm{3}{x}\:=\:\frac{\pi}{\mathrm{3}},\:\frac{\mathrm{4}\pi}{\mathrm{3}},\:\frac{\mathrm{7}\pi}{\mathrm{3}}\: \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:=\:\mathrm{tan}\:\frac{\pi}{\mathrm{3}}=\mathrm{tan}\:\frac{\mathrm{4}\pi}{\mathrm{3}}=\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{3}}\:=\:\sqrt{\mathrm{3}}…
Question Number 161698 by cortano last updated on 21/Dec/21 $$\:\:\mathrm{sin}\:\left({x}+{y}\right)=\mathrm{sin}\:{x}+\mathrm{sin}\:{y} \\ $$ Commented by Rasheed.Sindhi last updated on 21/Dec/21 $$\left({x},\mathrm{0}\right),\left(\mathrm{0},{y}\right),\left(\mathrm{0},\mathrm{0}\right)\:{are}\:{obvious}\:{solutions} \\ $$ Commented by cortano…
Question Number 161664 by cortano last updated on 21/Dec/21 $$\:\:\mathrm{5sec}\:\alpha\:−\mathrm{4}\:\mathrm{tan}\:\alpha\:=\:\mathrm{3cosec}\:\alpha \\ $$$$\:\:\frac{\mathrm{3cot}\:\alpha}{\mathrm{5}\:\mathrm{tan}\:\alpha−\mathrm{4}\:\mathrm{sec}\:\alpha}\:=?\: \\ $$ Answered by som(math1967) last updated on 21/Dec/21 $$\left(\mathrm{5}{sec}\alpha−\mathrm{4}{tan}\alpha\right)^{\mathrm{2}} =\mathrm{9}{cosec}^{\mathrm{2}} \alpha \\…
Question Number 30528 by abdo imad last updated on 22/Feb/18 $${simplify}\:\:{A}=\:{arctan}\left(\frac{{sinx}}{\mathrm{1}−{cosx}}\right)\:. \\ $$ Commented by abdo imad last updated on 23/Feb/18 $${we}\:{have}\:{A}={arctan}\left(\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\right) \\ $$$${A}={arctan}\left({cotan}\left(\frac{{x}}{\mathrm{2}}\right)\right)={arctan}\left(\:\frac{\mathrm{1}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\right)…
Question Number 30455 by ajfour last updated on 22/Feb/18 Commented by mrW2 last updated on 25/Feb/18 $${p}=\frac{{ac}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}}…
Question Number 95903 by 675480065 last updated on 28/May/20 Commented by PRITHWISH SEN 2 last updated on 28/May/20 $$\because\:\mathrm{x}>\mathrm{0} \\ $$$$\therefore\:\mathrm{x}^{\mathrm{2}} >\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}+\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{x}^{\mathrm{2}} <\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}<\mathrm{1}…
Question Number 161396 by abdullah_ff last updated on 17/Dec/21 $$\mathrm{If}\:{y}\:=\:{cos}^{\mathrm{2}} {x}\:+\:{sin}^{\mathrm{4}} {x}\:\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}, \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\leq\:{y}\:\leq\:\mathrm{1} \\ $$ Commented by cortano last updated on 17/Dec/21…