Question Number 147643 by bobhans last updated on 22/Jul/21 $$\:\mathrm{tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right)=\mathrm{tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}} \\ $$ Answered by liberty last updated on 22/Jul/21 $$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}=−\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\left[\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right]=−\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)=−\mathrm{3}\left(\frac{\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}{\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\right) \\…
Question Number 147581 by EDWIN88 last updated on 22/Jul/21 $$\:\:\mathrm{2sin}\:\mathrm{2x}\:−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{7cos}\:\mathrm{2x}\: \\ $$$$\:\frac{\pi}{\mathrm{2}}<\mathrm{x}<\pi\:\Rightarrow\:\mathrm{sin}\:\mathrm{2x}\:=? \\ $$ Answered by iloveisrael last updated on 22/Jul/21 $$\:\mathrm{If}\:\mathrm{2sin}\:\mathrm{2x}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{7cos}\:\mathrm{2x}\: \\…
Question Number 81983 by oyemi kemewari last updated on 17/Feb/20 Commented by jagoll last updated on 17/Feb/20 $${L}−{D}\:{or}\:{L}.{D}? \\ $$ Commented by oyemi kemewari last…
Question Number 81963 by jagoll last updated on 17/Feb/20 $${if}\:\mathrm{tan}\:\left({x}\right)+\mathrm{sec}\:\left({x}\right)\:=\:\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${find}\:\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\:=\: \\ $$ Commented by john santu last updated on 17/Feb/20 $${let}\:\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\:=\:{t} \\ $$$$\left({i}\right)\:\left\{\mathrm{tan}\:\left({x}\right)+\mathrm{sec}\:\left({x}\right)\right\}×\left\{\mathrm{cot}\:\left({x}\right)+\mathrm{cosec}\:\left({x}\right)\right\}=\:\frac{\mathrm{7}{t}}{\mathrm{8}}…
Question Number 16430 by Tinkutara last updated on 22/Jun/17 $$\mathrm{In}\:\Delta{ABC}\:\mathrm{with}\:\mathrm{usual}\:\mathrm{notation} \\ $$$$\frac{{r}_{\mathrm{1}} }{{bc}}\:+\:\frac{{r}_{\mathrm{2}} }{{ca}}\:+\:\frac{{r}_{\mathrm{3}} }{{ab}}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{{R}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{{r}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{{r}}\:+\:\frac{\mathrm{1}}{{R}} \\ $$…
Question Number 16392 by ajfour last updated on 21/Jun/17 Answered by Tinkutara last updated on 21/Jun/17 Commented by Tinkutara last updated on 21/Jun/17 $$\angle{AMI}\:=\:\mathrm{90}°,\:\angle{MAI}\:=\:\frac{{A}}{\mathrm{2}},\:{s}\:=\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}}, \\…
Question Number 16360 by ajfour last updated on 21/Jun/17 Answered by ajfour last updated on 21/Jun/17 $$\:{To}\:{find}\:{x}={CD},\:\:{y}={BD},\:{z}={AD} \\ $$$${BE}\:{is}\:{drawn}\:\bot\:{to}\:{AD},\:{AF}\:{is} \\ $$$${drawn}\:\bot\:{to}\:{AD}\:{produced}. \\ $$$$\:\bigtriangleup{BDE}\:\sim\:\bigtriangleup{ADF}\:\:\left({right}\:\angle\:{and}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:{vertically}\:{opposite}\:{angles}\right)…
Question Number 16358 by Tinkutara last updated on 21/Jun/17 $$\mathrm{In}\:\mathrm{any}\:\mathrm{triangle}\:{ABC},\:{a}\:\mathrm{cot}\:{A}\:+\:{b}\:\mathrm{cot}\:{B} \\ $$$$+\:{c}\:\mathrm{cot}\:{C}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{1}\right)\:{r}\:+\:{R} \\ $$$$\left(\mathrm{2}\right)\:{r}\:−\:{R} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}\left({r}\:+\:{R}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\left({r}\:−\:{R}\right) \\ $$ Commented by b.e.h.i.8.3.4.1.7@gmail.com…
Question Number 16359 by Tinkutara last updated on 21/Jun/17 $$\mathrm{In}\:\mathrm{a}\:\Delta{ABC}\:\mathrm{if}\:\frac{{s}\:−\:{a}}{{a}\:−\:{b}}\:=\:\frac{{s}\:−\:{c}}{{b}\:−\:{c}}\:,\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:{r}_{\mathrm{1}} ,\:{r}_{\mathrm{2}} ,\:{r}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Here}\:{r}_{\mathrm{1}} ,\:{r}_{\mathrm{2}} \:\mathrm{and}\:{r}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{exradii} \\ $$$$\mathrm{opposite}\:\mathrm{to}\:\mathrm{angles}\:{A},\:{B}\:\mathrm{and}\:{C}\:\mathrm{respectively}. \\ $$ Answered…
Question Number 16354 by Tinkutara last updated on 21/Jun/17 Answered by myintkhaing last updated on 21/Jun/17 $$\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}=\mathrm{2}−\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}}…