Question Number 11301 by tawa last updated on 19/Mar/17 $$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cyclic}\:\mathrm{quad}\:\mathrm{and}\:\mathrm{the}\:\mathrm{diagonal}\:\mathrm{AC}\:\mathrm{and}\:\mathrm{BD}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{H}.\: \\ $$$$\bigtriangleup\mathrm{DAC}\:=\:\mathrm{41}°\:\mathrm{and}\:\bigtriangleup\mathrm{AHB}\:=\:\mathrm{70}°.\:\mathrm{calulate}\:\:\bigtriangleup\mathrm{ACB} \\ $$ Commented by sandy_suhendra last updated on 20/Mar/17 $$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\angle\mathrm{DAC},\:\angle\mathrm{AHB}\:\mathrm{and}\:\angle\mathrm{ACB}? \\ $$ Answered…
Question Number 76790 by john santu last updated on 30/Dec/19 $${calculate}\:\frac{{sin}\mathrm{72}^{{o}} +{sin}\mathrm{148}^{{o}} +{sin}\mathrm{140}^{{o}} }{{sin}\mathrm{36}^{{o}} ×{sin}\mathrm{74}^{{o}} ×{sin}\mathrm{70}^{{o}} }. \\ $$ Commented by benjo 1/2 santuyy last…
Question Number 76791 by john santu last updated on 30/Dec/19 $${calculate}\:\frac{{sin}\mathrm{44}^{{o}} +{sin}\mathrm{66}^{{o}} +{sin}\mathrm{70}^{{o}} }{{cos}\mathrm{22}^{{o}} ×{cos}\mathrm{33}^{{o}} ×{cos}\mathrm{35}^{{o}} }. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 142168 by iloveisrael last updated on 27/May/21 Commented by liberty last updated on 27/May/21 $$=\:\mathrm{19} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 142167 by iloveisrael last updated on 27/May/21 $$\:\:\:{Find}\:{max}\:\&\:{min}\:{of}\:{function} \\ $$$$\:\:{f}\left({x}\right)=\mathrm{4tan}\:{x}+\mathrm{3cot}\:{x} \\ $$ Answered by MJS_new last updated on 27/May/21 $$−\infty<\mathrm{tan}\:{x}\:<+\infty\:\Rightarrow\:\mathrm{no}\:\mathrm{absolute}\:\mathrm{min},\:\mathrm{max} \\ $$$${g}\left({t}\right)=\mathrm{4}{t}+\frac{\mathrm{3}}{{t}} \\…
Question Number 76495 by mathocean1 last updated on 27/Dec/19 $$\mathrm{how}\:\mathrm{that} \\ $$$$\mathrm{tan2}{x}=\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}} \\ $$ Commented by mathmax by abdo last updated on 27/Dec/19 $${due}\:{to}\:{formula}\:\:{tan}\left(\mathrm{2}{x}\right)={tan}\left({x}+{x}\right)\:=\frac{{tanx}+{tanx}}{\mathrm{1}−{tanx}.{tanx}}…
Question Number 10955 by Joel576 last updated on 04/Mar/17 $$\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…