Question Number 28608 by abdo imad last updated on 27/Jan/18 $${transform}\:{tanp}−{tanq}\:{then}\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{cos}\left({k}\theta\right){cos}\left(\left({k}+\mathrm{1}\right)\theta\right.}\:.\:\:\theta\in{R}. \\ $$ Answered by Tinkutara last updated on 28/Jan/18 $$\frac{\mathrm{1}}{\mathrm{cos}\:{k}\theta\mathrm{cos}\:\left({k}+\mathrm{1}\right)\theta}…
Question Number 28583 by tawa tawa last updated on 27/Jan/18 Commented by tawa tawa last updated on 27/Jan/18 $$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{above}:\:\:\:\:\mathrm{AB}\:\mathrm{is}\:\mathrm{a}\:\mathrm{chord}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{10cm}.\:\mathrm{M}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{mid}\:\mathrm{point}\:\mathrm{of}\:\mathrm{AB}.\:\mathrm{and}\:\:\mathrm{NM}\:\bot\:\mathrm{AB} \\ $$ Answered by…
Question Number 159641 by tounghoungko last updated on 19/Nov/21 $${minimum}\:{value}\:{of}\:{f}\left({x}\right)=\mathrm{256}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)+\mathrm{324}\:\mathrm{cosec}\:^{\mathrm{2}} \left({x}\right) \\ $$$$\:\forall{x}\in\:\mathbb{R}\: \\ $$ Commented by gsk2684 last updated on 19/Nov/21 $${minimum}\:{value}\:{of}\:{f}\left({x}\right)=\mathrm{256}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}\right)+\mathrm{324}\:\mathrm{cosec}\:^{\mathrm{2}}…
Question Number 159642 by tounghoungko last updated on 19/Nov/21 $${minimum}\:{value}\:{of}\:{function}\: \\ $$$$\:\:\:{f}\left({x}\right)=\sqrt{\left(\mathrm{3sin}\:{x}−\mathrm{4cos}\:{x}−\mathrm{10}\right)\left(\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}−\mathrm{10}\right)} \\ $$ Commented by bobhans last updated on 19/Nov/21 $$\:\mathrm{g}\left(\mathrm{x}\right)=\left(\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)−\mathrm{4cos}\:\mathrm{x}\right)\left(\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)+\mathrm{4cos}\:\mathrm{x}\right) \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\left(\mathrm{3sin}\:\mathrm{x}−\mathrm{10}\right)^{\mathrm{2}} −\mathrm{16cos}\:^{\mathrm{2}}…
Question Number 159635 by cortano last updated on 19/Nov/21 Commented by cortano last updated on 19/Nov/21 $$\:\mathrm{tan}\:\alpha\:=\:\frac{{x}}{\mathrm{4}}\:;\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{3}{x}}{\mathrm{12}\sqrt{\mathrm{3}}}\:=\:\frac{{x}}{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}\:\alpha\: \\ $$$$\Rightarrow\alpha+\beta+\mathrm{105}°=\mathrm{180}° \\ $$$$\Rightarrow\alpha+\beta\:=\:\mathrm{75}° \\ $$$$\Rightarrow\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\:…
Question Number 94071 by oustmuchiya@gmail.com last updated on 16/May/20 Commented by hknkrc46 last updated on 16/May/20 $$\left(\mathrm{6}\right)\:\mathrm{sin}\:\left(\mathrm{180}^{\circ} +\mathrm{A}\right)=−\mathrm{sin}\:\mathrm{A} \\ $$$$\mathrm{cos}\:\left(\mathrm{90}^{\circ} −\mathrm{A}\right)=\mathrm{sin}\:\mathrm{A} \\ $$$$\mathrm{tan}\:\left(\mathrm{270}^{\circ} −\mathrm{A}\right)=\mathrm{cot}\:\mathrm{A} \\…
Question Number 28515 by beh.i83417@gmail.com last updated on 26/Jan/18 Answered by mrW2 last updated on 26/Jan/18 $$\mathrm{tan}\:\mathrm{20}−\mathrm{tan}\:\mathrm{40}+\mathrm{tan}\:\mathrm{80} \\ $$$$=\mathrm{tan}\:\mathrm{20}−\mathrm{tan}\:\left(\mathrm{60}−\mathrm{20}\right)+\mathrm{tan}\:\left(\mathrm{60}+\mathrm{20}\right) \\ $$$$=\mathrm{tan}\:\mathrm{20}−\frac{\sqrt{\mathrm{3}}−\mathrm{tan}\:\mathrm{20}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{20}}+\frac{\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{20}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{20}} \\ $$$$=\mathrm{tan}\:\mathrm{20}+\frac{\mathrm{tan}\:\mathrm{20}−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{20}}+\frac{\mathrm{tan}\:\mathrm{20}+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{20}} \\ $$$$=\mathrm{tan}\:\mathrm{20}+\frac{\mathrm{8tan}\:\mathrm{20}}{\mathrm{1}−\mathrm{3tan}^{\mathrm{2}}…
Question Number 93952 by oustmuchiya@gmail.com last updated on 16/May/20 $${Express}\:{the}\:{following}\:{as}\:{functions} \\ $$$${of}\:\boldsymbol{{A}}: \\ $$$$\left(\mathrm{i}\right)\:{sec}\left({A}−\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:\:\left(\mathrm{ii}\right)\:{cosec}\left({A}−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{iii}\right)\:{tan}\left({A}−\frac{\mathrm{3}\pi_{} }{\mathrm{2}}\right)\:\:\left(\mathrm{iv}\right)\:{cos}\left(\mathrm{720}°+{A}\right) \\ $$$$\left(\mathrm{v}\right)\:{tan}\:\left({A}+\pi\right) \\ $$ Answered by Kunal12588 last…
Question Number 93940 by seedhamaieng@gmail.com last updated on 16/May/20 Commented by Tony Lin last updated on 16/May/20 $${cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}={tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}} \\ $$$${tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}+{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\…
Question Number 159466 by abdullah_ff last updated on 17/Nov/21 $$\mathrm{if}\:{tanA}\:=\:\frac{{a}}{{b}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{sinA}\:=\:\pm\:\frac{{a}}{\:\sqrt{{a}^{ } \:+\:{b}^{ } }} \\ $$$$\mathrm{please}\:\mathrm{help}.. \\ $$ Answered by Ar Brandon last…