Question Number 10955 by Joel576 last updated on 04/Mar/17 $$\mathrm{If}\:\:\frac{\mathrm{cos}\:\theta}{\mathrm{1}\:−\:\mathrm{sin}\:\theta}\:=\:{a}\:\:\:\:\:\:\:\:\:\:\:{a}\:\neq\:\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2}{k}\pi \\ $$$$\mathrm{So},\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:=\:… \\ $$$$\left(\mathrm{A}\right)\:\:\frac{{a}}{{a}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\:\frac{{a}\:+\:\mathrm{1}}{{a}\:−\:\mathrm{1}} \\ $$$$\left(\mathrm{B}\right)\:\:\frac{\mathrm{1}}{{a}\:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{E}\right)\:\:\frac{{a}}{{a}\:−\:\mathrm{1}} \\ $$$$\left(\mathrm{C}\right)\:\:\frac{{a}\:−\:\mathrm{1}}{{a}\:+\:\mathrm{1}} \\ $$ Answered by ajfour last updated…
Question Number 10948 by rish@bh last updated on 03/Mar/17 $$\mathrm{If}\:\mathrm{cos}^{−\mathrm{1}} \frac{{x}}{{a}}+\mathrm{cos}^{−\mathrm{1}} \frac{{y}}{{b}}=\alpha \\ $$$${prove}\: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{xy}}{{ab}}\mathrm{cos}\:\alpha+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{sin}^{\mathrm{2}} \alpha \\ $$ Answered by…
In-a-triangle-ABC-prove-the-following-a-b-c-2-a-2-b-2-c-2-cot-A-2-cot-B-2-cot-C-2-cot-A-cot-B-cot-C-
Question Number 10947 by rish@bh last updated on 03/Mar/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following} \\ $$$$\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{cot}\:\frac{{A}}{\mathrm{2}}+\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}}{\mathrm{cot}\:{A}+\mathrm{cot}\:{B}+\mathrm{cot}\:{C}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76477 by mathocean1 last updated on 27/Dec/19 $$\mathrm{Hello}\:\mathrm{Solve}\:\mathrm{in}\: \\ $$$$\left.\right]−\pi;\pi\left[\right. \\ $$$$\frac{\mathrm{2sin}{x}}{{cosx}−{sinx}}\geqslant\mathrm{0} \\ $$$$\mathrm{result}\:\mathrm{should}\:\mathrm{be}\:\mathrm{in}\:\mathrm{radian} \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:… \\ $$ Answered by Kunal12588 last updated…
Question Number 142010 by jlewis last updated on 25/May/21 $$\mathrm{find}\:\mathrm{x}\:\mathrm{2cosh}\:\mathrm{2x}\:+\mathrm{10sinh}\:\mathrm{2x}=\mathrm{5} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10917 by okhema last updated on 02/Mar/17 $${hence}\:{show}\:{that} \\ $$$$\left.{i}\right)\frac{\mathrm{1}−{cos}\mathrm{4}\theta}{{sin}\mathrm{4}\theta}={tan}\mathrm{2}\theta \\ $$$$\left.{ii}\right)\frac{\mathrm{1}−{cos}\mathrm{6}\theta}{{sin}\mathrm{6}\theta}={tan}\mathrm{3}\theta \\ $$ Answered by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{we}\:\mathrm{use}\:\:\:\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\:\Rightarrow\:\mathrm{2sin}^{\mathrm{2}}…
Question Number 10916 by okhema last updated on 02/Mar/17 $${find}\:{all}\:{possible}\:{values}\:{of}\:{cos}\theta\:{such} \\ $$$${that}\:\mathrm{2}{cot}^{\mathrm{2}} \theta+{cos}\theta=\mathrm{0} \\ $$ Answered by sandy_suhendra last updated on 02/Mar/17 $$\frac{\mathrm{2cos}^{\mathrm{2}} \theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:+\:\mathrm{cos}\theta\:=\mathrm{0}…
Question Number 10906 by niraj last updated on 01/Mar/17 $${Q}\:.\:\:{smallest}\:{positive}\:{x}\:{satisfying}\:{the}\:{equation} \\ $$$${sin}\mathrm{3}{x}+\mathrm{3}{cosx}=\mathrm{2}{sin}\mathrm{2}{x}\left({sinx}+{cosx}\right)\:,\:{is} \\ $$ Commented by niraj last updated on 02/Mar/17 $${sir}\:{answer}\:{please} \\ $$ Answered…
Question Number 76439 by mathocean1 last updated on 27/Dec/19 $$\mathrm{solve}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\left[\:\right.\right. \\ $$$$\mathrm{1}+\mathrm{2sin3}{x}\leqslant\mathrm{0} \\ $$ Answered by john santu last updated on 27/Dec/19 $${sin}\mathrm{3}{x}=−\frac{\mathrm{1}}{\mathrm{2}}.\rightarrow{x}=\left\{\mathrm{10}^{{o}} ,\mathrm{50}^{{o}} ,\mathrm{130}^{{o}}…
Question Number 76434 by benjo last updated on 27/Dec/19 $$\mathrm{if}\:\mathrm{sec}\:\mathrm{x}\:−\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\mathrm{sin}\:\mathrm{x}. \\ $$ Answered by john santu last updated on 27/Dec/19 $${using}\::\:\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{tan}\:^{\mathrm{2}} {x}\:=\mathrm{1}…