Question Number 62761 by lalitchand last updated on 25/Jun/19 $$\mathrm{if}\:\mathrm{cos}\theta=\frac{\mathrm{cosA}.\mathrm{cosB}}{\mathrm{1}−\mathrm{cosA}.\mathrm{cosB}}\:\:\:\mathrm{prove}\:\mathrm{that}\:\mathrm{Tan}\frac{\theta}{\mathrm{2}}=\mathrm{Tan}\frac{\mathrm{A}}{\mathrm{2}}.\mathrm{Cot}\frac{\mathrm{B}}{\mathrm{2}} \\ $$ Commented by Prithwish sen last updated on 25/Jun/19 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{cos}\theta\:=\:\frac{\mathrm{cosA}−\mathrm{cosB}}{\mathrm{1}−\mathrm{cosAcosB}} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}\:=\:\frac{\mathrm{1}−\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosAcosB}}{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}−\mathrm{cosAcosB}}…
Question Number 62698 by Ankit0512 last updated on 24/Jun/19 $${Lines}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{10}=\mathrm{0}\:{and}\:\mathrm{5}{x}−\mathrm{12}{y}−\mathrm{40}=\mathrm{0} \\ $$$${touch}\:{circle}\:{C}_{\mathrm{1}} \:{of}\:{diameter}\:\mathrm{6}.\:{If}\:{the}\: \\ $$$${center}\:{of}\:{C}_{\mathrm{1}} \:{lies}\:{in}\:{the}\:{Ist}\:{quadrant}, \\ $$$${find}\:{the}\:{equation}\:{of}\:{circle}\:{C}_{\mathrm{2}} \:{which}\:{is}\: \\ $$$${concentric}\:{with}\:{C}_{\mathrm{1}\:} \:{and}\:{cuts}\:{intercept} \\ $$$${of}\:{length}\:\mathrm{8}\:{on}\:{these}\:{lines}. \\…
Question Number 128178 by bemath last updated on 05/Jan/21 $$\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{any}\: \\ $$$$\mathrm{quadrilateral}\:\mathrm{ABCD}\:\mathrm{is}\: \\ $$$$\:\sqrt{\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)\left(\mathrm{s}−\mathrm{d}\right)−\mathrm{abcd}\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{A}+\mathrm{C}}{\mathrm{2}}\right)}\:. \\ $$ Answered by mr W last updated on 05/Jan/21…
Question Number 62595 by Pranay last updated on 23/Jun/19 $${what}\:{is}\:\mathrm{2}{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}} \theta\:? \\ $$ Commented by ajfour last updated on 23/Jun/19 $${same}\:{as}\:\:\mathrm{2}×\frac{{p}^{\mathrm{2}} }{{h}^{\mathrm{2}} }×\frac{{b}^{\mathrm{2}} }{{h}^{\mathrm{2}}…
Question Number 128130 by physicstutes last updated on 04/Jan/21 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\mathrm{sin}\left(\mathrm{2}{x}\:+\:\mathrm{2}{h}\right)−\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{2cos}\left(\mathrm{2}{x}\:+{h}\right)\mathrm{sin}\:{h} \\ $$ Answered by Olaf last updated on 04/Jan/21 $$\mathrm{sin}{a}−\mathrm{sin}{b}\:=\:\mathrm{2sin}\left(\frac{{a}−{b}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right)\:\left(\mathrm{1}\right) \\ $$$${a}\:=\:\mathrm{2}{x}+\mathrm{2}{h}\:\mathrm{and}\:{b}\:=\:\mathrm{2}{x} \\…
Question Number 128094 by TITA last updated on 04/Jan/21 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\:\mathrm{in}\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$$$\mathrm{sin}\:\mathrm{2x}=\mathrm{sec}\:\mathrm{x} \\ $$ Answered by MJS_new last updated on 04/Jan/21 $$\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x} \\ $$$$\mathrm{sin}\:{x}\:={s} \\…
Question Number 128068 by benjo_mathlover last updated on 04/Jan/21 $$\:\mathrm{Given}\:\mathrm{25x}^{\mathrm{2}} −\mathrm{30x}+\mathrm{7}=\mathrm{0}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\mathrm{are}\:\mathrm{cos}\:\alpha\:\mathrm{and}\:\mathrm{cos}\:\beta.\:\mathrm{If}\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta>\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\right).\mathrm{tan}\:\left(\frac{\alpha−\beta}{\mathrm{2}}\right)\:=? \\ $$ Answered by bemath last updated on 04/Jan/21 $$\:\mathrm{let}\:\frac{\alpha+\beta}{\mathrm{2}}=\mathrm{x}\:\wedge\:\frac{\alpha−\beta}{\mathrm{2}}=\mathrm{y}…
Question Number 128010 by Study last updated on 03/Jan/21 Commented by Study last updated on 03/Jan/21 $${help}\:{me}\:{and}\:{show}\:{practice} \\ $$ Commented by som(math1967) last updated on…
Question Number 62431 by rajesh4661kumar@gamil.com last updated on 21/Jun/19 Answered by tanmay last updated on 21/Jun/19 $${asin}^{\mathrm{2}} \theta+{bcos}^{\mathrm{2}} \theta={c} \\ $$$${asin}^{\mathrm{2}} \theta+{b}\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right)={c} \\ $$$${sin}^{\mathrm{2}}…
Question Number 127944 by liberty last updated on 03/Jan/21 $$\:\mathrm{If}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{then}\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:=? \\ $$ Answered by mr W last updated on 03/Jan/21 $${p}_{{n}} =\mathrm{sin}^{{n}} \:{x}+\mathrm{cos}^{{n}}…