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Category: Trigonometry

Question-46283

Question Number 46283 by somil last updated on 23/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18 $${PQ}=\mathrm{12}\:\:{QR}=\mathrm{5}\:\:{PR}=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:=\mathrm{13} \\ $$$$ \\ $$$$ \\ $$$$\angle{DQP}=\theta\:\:\angle{DQR}=\mathrm{90}−\theta…

Find-the-positive-integer-n-such-that-tan-1-1-3-tan-1-1-4-tan-1-1-5-tan-1-1-n-pi-4-

Question Number 111725 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)=\frac{\pi}{\mathrm{4}} \\ $$ Answered by $@y@m last updated…

Question-111426

Question Number 111426 by weltr last updated on 03/Sep/20 Answered by Dwaipayan Shikari last updated on 03/Sep/20 $$\mathrm{4}\left({sin}^{\mathrm{6}} \alpha+{cos}^{\mathrm{6}} \alpha\right)−\mathrm{3}{cos}^{\mathrm{2}} \mathrm{2}\alpha \\ $$$$=\mathrm{4}\left({sin}^{\mathrm{2}} \alpha+{cos}^{\mathrm{2}} \alpha\right)^{\mathrm{3}}…

Compute-cos-12-

Question Number 111391 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{Compute}\:\mathrm{cos}\frac{\Pi}{\mathrm{12}} \\ $$ Answered by bemath last updated on 03/Sep/20 $$\frac{\pi}{\mathrm{6}}\:=\:\mathrm{2}×\frac{\pi}{\mathrm{12}}\:\Rightarrow\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:=\:\mathrm{cos}\:\left(\mathrm{2}.\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:=\:\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)−\mathrm{1}…

Question-111382

Question Number 111382 by john santu last updated on 03/Sep/20 Commented by kaivan.ahmadi last updated on 03/Sep/20 $${sec}^{\mathrm{2}} {x}=\mathrm{12}+{secx}\Rightarrow{sec}^{\mathrm{2}} {x}−{secx}−\mathrm{12}=\mathrm{0}\Rightarrow \\ $$$$\left({secx}−\mathrm{4}\right)\left({secx}+\mathrm{3}\right)=\mathrm{0}\Rightarrow{secx}=\mathrm{4},\:{secx}=−\mathrm{3} \\ $$$${by}\:{hypothesis}\:{secx}>\mathrm{0}\:{so}\:{we}\:{have}\:{secx}=\mathrm{4} \\…

Question-45794

Question Number 45794 by Tawa1 last updated on 16/Oct/18 Answered by MJS last updated on 16/Oct/18 $${a}\ast{c}={b}\:\Rightarrow\:{b}\ast{c}^{−\mathrm{1}} ={a}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${b}\ast{c}={d}\:\Rightarrow\:{d}\ast{c}^{−\mathrm{1}} ={b}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${c}\ast{c}={a}\:\Rightarrow\:{a}\ast{c}^{−\mathrm{1}}…