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…
Question Number 222025 by fantastic last updated on 15/Jun/25 $$\left(\frac{\mathrm{5cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{3}}+\mathrm{4sec}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}−\mathrm{tan}\:^{\mathrm{2}} \frac{\pi}{\mathrm{4}}}{\mathrm{sin}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}+\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{6}}}\right)=?? \\ $$$$\left[{easy}\:{mode}\right] \\ $$ Answered by MrGaster last updated on…
Question Number 222022 by fantastic last updated on 15/Jun/25 $${If}\:\angle{P}+\angle{Q}\:=\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{that} \\ $$$$\sqrt{\frac{\mathrm{sin}\:{P}}{\mathrm{cos}\:{Q}}−\mathrm{sin}\:{P}\mathrm{cos}\:{Q}}=\mathrm{cos}\:{P} \\ $$ Commented by MathematicalUser2357 last updated on 17/Jun/25 $${You}\:{mean}\:{If}\:\angle{P}+\angle{Q}\:=\mathrm{222022}^{\mathrm{0}} \:{then}\:{prove}\:{that} \\…
Question Number 221686 by Tawa11 last updated on 09/Jun/25 Commented by AlagaIbile last updated on 09/Jun/25 $$\:{Let}\:\mathrm{tan}\:\boldsymbol{{x}}\:=\:\boldsymbol{{y}},\:\mathrm{cos}\:\boldsymbol{{x}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}} \\ $$$$\Rightarrow\:\mathrm{cos}^{\mathrm{2}\:} \left[\mathrm{cos}^{-\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{y}}^{\mathrm{2}} \:+\:\mathrm{1}}}\right] \\ $$$$\Rightarrow\:\left[\mathrm{cos}\:\left(\mathrm{cos}^{-\mathrm{1}}…
Question Number 221578 by efronzo1 last updated on 08/Jun/25 $$\:\:\: \\ $$ Answered by gregori last updated on 08/Jun/25 $$\:\:\underbrace{\:}\: \\ $$ Terms of Service…
Question Number 221407 by fantastic last updated on 03/Jun/25 $${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221411 by ajfour last updated on 03/Jun/25 Answered by mr W last updated on 04/Jun/25 $${r}=\frac{\theta}{\mathrm{2}\:\mathrm{sin}\:\theta}=\frac{\pi}{\mathrm{6}} \\ $$ Commented by mr W last…
Question Number 220857 by fantastic last updated on 20/May/25 $${Prove}\:{that}\:\mathrm{tan}\:\mathrm{20}^{\mathrm{0}} \mathrm{tan40}^{\mathrm{0}} \:\mathrm{tan}\:\mathrm{80}^{\mathrm{0}} =\mathrm{tan}\:\mathrm{60}^{\mathrm{0}} \\ $$ Answered by fantastic last updated on 20/May/25 $$\mathrm{tan}\:\mathrm{20}^{\mathrm{0}} \mathrm{tan40}^{\mathrm{0}} \:\mathrm{tan}\:\mathrm{80}^{\mathrm{0}}…
Question Number 220855 by fantastic last updated on 20/May/25 $${If}\:\:{b}\:\mathrm{cos}\left(\theta+\mathrm{120}^{\mathrm{0}} \right)={c}\:\mathrm{cos}\:\left(\theta+\mathrm{240}^{\mathrm{0}} \right)\:{then}\:{prove}\:{that} \\ $$$${b}−{c}=−\left({b}+{c}\right)\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta \\ $$ Answered by golsendro last updated on 20/May/25 $$\:\mathrm{b}\:\mathrm{cos}\:\left(\mathrm{180}−\left(\mathrm{60}−\theta\right)\right)\:=\:\mathrm{c}\:\mathrm{cos}\:\left(\mathrm{360}−\left(\mathrm{120}−\theta\right)\right) \\…
Question Number 220712 by Nicholas666 last updated on 18/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}: \\ $$$$\:\:\frac{\mathrm{2}\:{tan}\:\mathrm{2}{A}\:+\:{tan}\:{A}}{\mathrm{4}\:{tan}\:\mathrm{3}{A}\:−\:{tan}\:\mathrm{2}{A}}\:=\:{sin}^{\mathrm{2}} \:{A} \\ $$$$\: \\ $$ Terms of Service Privacy Policy Contact:…