Question Number 8314 by tawakalitu last updated on 07/Oct/16 Commented by Tinku Tara last updated on 08/Oct/16 $$\mathrm{For}\:\mathrm{othercases}\:\mathrm{can}\:\mathrm{u}\:\mathrm{please}\:\mathrm{email} \\ $$$$\mathrm{picture}\:\mathrm{to}\:\mathrm{us}\:\mathrm{at}\:\mathrm{infoattinkutara}.\mathrm{com} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{will}\:\mathrm{troubleshoot}\:\mathrm{the}\:\mathrm{issue}. \\ $$$$\mathrm{Alternatively}\:\mathrm{try}\:\mathrm{to}\:\mathrm{post}\:\mathrm{image}\:\mathrm{as}\:\mathrm{a}\:\mathrm{new} \\…
Question Number 8310 by lepan last updated on 07/Oct/16 $${Given}\:{that}\:{cosecA}+{cotA}=\mathrm{3}\:{evaluate} \\ $$$${cosecA}−{cotA}\:{and}\:{cosA}. \\ $$$$ \\ $$ Answered by prakash jain last updated on 08/Oct/16 $$\mathrm{1}+\mathrm{cot}^{\mathrm{2}}…
Question Number 8306 by lepan last updated on 07/Oct/16 $${If}\:\mathrm{270}°<{x}<\mathrm{360}°,\:{simplify} \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cosx}}}. \\ $$ Commented by sou1618 last updated on 07/Oct/16 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cosx}}}\:\left(\mathrm{270}°<{x}<\mathrm{360}°\right) \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \alpha=\mathrm{1}+{cos}\mathrm{2}\alpha…
Question Number 8297 by lepan last updated on 06/Oct/16 $$\underset{} {{B}y}\:{expessing}\:{each}\:{side}\:{of}\:{the} \\ $$$${equation}\:{in}\:{terms}\:{of}\:{tanA}\:,{or}\: \\ $$$${otherwise}\:{show}\:{that} \\ $$$$\frac{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}+\mathrm{1}}{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}−\mathrm{1}}=\frac{{tan}\left(\mathrm{45}°+{A}\right)}{{tanA}} \\ $$ Answered by Rasheed Soomro last updated…
Question Number 8289 by rhm last updated on 07/Oct/16 $${what}\:{is}\:{value} \\ $$$${sin}\:\mathrm{36}° \\ $$$${plese}\:{give}\:{me}\:{answer} \\ $$ Commented by rhm last updated on 07/Oct/16 $${sir}\:{answer} \\…
Question Number 8287 by lepan last updated on 06/Oct/16 $${Show}\:{that}\:{tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}. \\ $$ Answered by ridwan balatif last updated on 06/Oct/16 $$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{sin}\left(\alpha+\beta\right)}{\mathrm{cos}\left(\alpha+\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha}{\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{sin}\alpha\mathrm{cos}\beta+\mathrm{sin}\beta\mathrm{cos}\alpha\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}{\left(\mathrm{cos}\alpha\mathrm{cos}\beta−\mathrm{sin}\alpha\mathrm{sin}\beta\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\alpha\mathrm{cos}\beta}\right)}…
Question Number 73817 by Rio Michael last updated on 16/Nov/19 $${find}\:{the}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$${in}\:\:\mathrm{0}\:\leqslant\:\theta\:\leqslant\:\pi \\ $$$$\:\:{sin}\mathrm{2}\theta\:=\:{sec}\theta \\ $$ Answered by mr W last updated on 16/Nov/19…
Question Number 8273 by lepan last updated on 05/Oct/16 $${Express}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha\:{in}\:{the}\:{form}\: \\ $$$${Rsin}\left(\alpha+\beta\right)\:{where}\:{R}>\mathrm{0}\:{and}\:\mathrm{0}°<\beta<\mathrm{90}°. \\ $$$${Hence}\:{solve}\:{the}\:{equation}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha=\mathrm{2} \\ $$$${for}\:\mathrm{0}°<\alpha<\mathrm{270}°. \\ $$ Answered by prakash jain last updated on…
Question Number 8275 by lepan last updated on 05/Oct/16 $${Show}\:{that}\:{the}\:{followings} \\ $$$$\left({i}\right){sin}\left({a}+{b}\right)={sina}\:{cosb}\:+{cosa}\:{sinb} \\ $$$$\left({ii}\right){cos}\left({a}−{b}\right)={cosa}\:{cosb}\:+{sina}\:{sinb} \\ $$$$ \\ $$ Commented by sou1618 last updated on 05/Oct/16…
Question Number 8269 by tawakalitu last updated on 04/Oct/16 Commented by tawakalitu last updated on 04/Oct/16 $$\mathrm{Find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}.\:\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}. \\ $$ Answered by prakash jain last updated…