Question Number 155810 by zainaltanjung last updated on 05/Oct/21 $$\mathrm{Verify}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{in}\:\mathrm{Excercise}\:\mathrm{below} \\ $$$$\left.\mathrm{1}\right).\:\mathrm{cos}\:\theta\mathrm{sec}\:\theta=\mathrm{1} \\ $$$$\left.\mathrm{2}\right).\:\left(\mathrm{1}+\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\beta\right)=\mathrm{sin}\:^{\mathrm{2}} \beta \\ $$$$\left.\mathrm{3}\right).\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)=\mathrm{sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\left.\mathrm{4}\right).\:\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{cosec}\:\mathrm{t}}+\frac{\mathrm{cos}\:\mathrm{t}}{\mathrm{sec}\:\mathrm{t}}=\mathrm{1} \\ $$$$\left.\mathrm{5}\right).\:\frac{\mathrm{cosec}\:^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}}…
Question Number 24554 by tawa tawa last updated on 20/Nov/17 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{2q}}\right)\:+\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{q}}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$ Commented by mrW1 last updated on 21/Nov/17 $${that}'{s}\:{not}\:{true}! \\ $$$${let}'{s}\:{say}\:{p}={q}=\mathrm{1}…
Question Number 89958 by 20000193 last updated on 20/Apr/20 $${prove}\:{that}\:\left(\mathrm{1}+\mathrm{sin}\:{x}/\mathrm{1}+\mathrm{cos}\:\mathrm{3}\left(\mathrm{1sin}\:{x}/\mathrm{1}+\mathrm{cosec}\:{x}\right)={tanx}\right. \\ $$ Commented by john santu last updated on 20/Apr/20 $$\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{3}\left(\mathrm{1sin}\:{x}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosec}\:{x}\:}\right)}\:=\:\mathrm{tan}\:{x}? \\ $$ Terms of…
Question Number 89951 by john santu last updated on 20/Apr/20 $$\mathrm{log}\:_{\mathrm{2}} \:\left(\mathrm{sin}\:\left({x}+\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right)\:+\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{12}}\right)\right)=−\mathrm{1} \\ $$ Commented by jagoll last updated on 20/Apr/20 $$\Rightarrow\mathrm{sin}\:\left(\mathrm{x}+\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\:\mathrm{sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{12}}\right)−\mathrm{cos}\:\left(\mathrm{2x}+\frac{\mathrm{6}\pi}{\mathrm{12}}\right)\:=\:\mathrm{1}…
Question Number 155466 by cortano last updated on 01/Oct/21 $$\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\ldots+\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)=? \\ $$ Commented by VIDDD last updated on 01/Oct/21 $${can}\:{u}\:{show}\:{u}\:{solution} \\ $$…
Question Number 24355 by ahmetbak1r last updated on 16/Nov/17 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{4}+\sqrt{\mathrm{16}+\sqrt{\mathrm{256}…..}}}}=? \\ $$ Answered by ajfour last updated on 16/Nov/17 $${S}=\sqrt{\mathrm{2}^{\mathrm{0}} +\sqrt{\mathrm{2}^{\mathrm{2}} +\sqrt{\mathrm{2}^{\mathrm{4}} +\sqrt{\mathrm{2}^{\mathrm{8}} +\sqrt{..}}}}} \\…
Question Number 89848 by 9933039645 last updated on 19/Apr/20 $${Q}\mathrm{1}.\:{If}\:{sin}\theta=\frac{\mathrm{3}}{\mathrm{5}}\:{then}\:{find}\:{the}\:{value}\:{of}\:{tan}\theta+{cot}\theta \\ $$ Commented by mahdi last updated on 19/Apr/20 $$\pm\frac{\mathrm{25}}{\mathrm{12}} \\ $$$$\mathrm{sin}\theta=\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow\mid\mathrm{cos}\theta\mid=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{tan}\theta=\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}=\pm\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\mathrm{cot}\theta=\frac{\mathrm{1}}{\mathrm{tan}\theta}=\pm\frac{\mathrm{4}}{\mathrm{3}}…
Question Number 24294 by ajfour last updated on 15/Nov/17 $${Find}\:{value}\left({s}\right)\:{of}\:\boldsymbol{{x}}\:{if} \\ $$$$\:\:\mathrm{sin}\:\left[\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}} {x}\right)\right\}\right]=\mathrm{0}\:. \\ $$ Answered by mrW1 last updated on 15/Nov/17 $$\:\mathrm{sin}\:\left[\mathrm{2cos}^{−\mathrm{1}} \left\{\mathrm{cot}\:\left(\mathrm{2tan}^{−\mathrm{1}}…
Question Number 89662 by 974342176 last updated on 18/Apr/20 $${Given}\:{that}\:\mathrm{sin}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\mathrm{sin}\:{C}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{without}\:{u} \\ $$$${using}\:{calculator}\:{solve} \\ $$$$\left.{a}\right)\:\mathrm{tan}\:\left({A}+{C}\right) \\ $$$$\left.{b}\right)\:{cos}\left({A}−{C}\right) \\ $$ Answered by TANMAY PANACEA. last updated on…
Question Number 24098 by Sudipta Jana last updated on 12/Nov/17 $$\mathrm{sin}^{−\mathrm{1}} \frac{{ax}}{{c}}+\mathrm{sin}^{−\mathrm{1}} \frac{{bx}}{{c}}=\mathrm{sin}^{−\mathrm{1}} {x}\:\:\:\:\:\left[{When}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\right] \\ $$ Answered by 951172235v last updated on…