Category: Trigonometry

Question Number 161254 by cortano last updated on 15/Dec/21 $$\frac{4\sin \left(\frac{2\pi }{7}\right)+\sec \left(\frac{\pi }{14}\right)}{\cot \left(\frac{\pi }{7}\right)}=?$$ Commented by bobhans last updated on 15/Dec/21 $$\mathcal{P}=\frac{4\sin \left(\frac{2\pi }{7}\right)\cos \left(\frac{\pi }{14}\right)+1}{\cos \left(\frac{\pi }{14}\right)\left[\frac{\cos \left(\frac{\pi }{7}\right)}{2\sin \left(\frac{\pi }{14}\right)\cos \left(\frac{\pi }{14}\right)}\right]}$$$$\mathcal{P}=\frac{4\sin \left(\frac{2\pi }{7}\right)\sin \left(\frac{\pi }{7}\right)+2\sin \left(\frac{\pi }{14}\right)}{\cos \left(\frac{\pi }{7}\right)}$$$$\:\mathcal{P}\:=\:\frac{−\mathrm{2}\left(\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)\right)+\mathrm{2cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)} \…

Question Number 161181 by cortano last updated on 13/Dec/21 $$x={\cot }^{-1}\left(\sqrt{\cos \theta }\right)-{\tan }^{-1}\left(\sqrt{\cos \theta }\right)$$$$\sin x=?$$ Commented by MJS_new last updated on 13/Dec/21 $$\mathrm{sin}\:\left(\mathrm{arctan}\:{a}\:−\mathrm{arctan}\:{b}\right)\:=\frac{{a}−{b}}{\:\sqrt{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}}…

Question Number 30054 by rahul 19 last updated on 15/Feb/18 Answered by ajfour last updated on 16/Feb/18 $$let\sin x=s$$$$\mathrm{5}{s}^{\mathrm{2}} +\mathrm{4}{s}^{\mathrm{2}} \left(\mathrm{1}−{s}^{\mathrm{2}} \right)−\mathrm{4}\left(\mathrm{1}−\mathrm{2}{s}^{\mathrm{2}} \right)\:>\:\mathrm{0} \…

Question Number 95531 by Abdulrahman last updated on 25/May/20 $$\frac{\mathrm{tanx}\times \mathrm{ctg}2\mathrm{x}}{{\tan }^2\mathrm{x}-1}=?$$ Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\mathrm{put}\:\mathrm{cot2x}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2tan}\:\mathrm{x}} \…

Question Number 95524 by Abdulrahman last updated on 25/May/20 $$\frac{1}{\sin 10}-\frac{\sqrt{3}}{\cos 10}=?$$ Commented by PRITHWISH SEN 2 last updated on 25/May/20 $$\frac{\cos 10-\sqrt{3}\sin 10}{\sin 10\cos 10}=4\frac{(\frac{1}{2}\cos 10-\frac{\sqrt{3}}{2}\sin 10)}{2\sin 10\cos 10}$$$$=\mathrm{4}.\frac{\mathrm{sin}\:\mathrm{30}.\mathrm{cos}\:\mathrm{10}−\mathrm{cos}\:\mathrm{30}.\mathrm{sin}\:\mathrm{10}}{\mathrm{sin}\:\mathrm{20}}\:=\mathrm{4}.\frac{\mathrm{sin}\:\left(\mathrm{30}−\mathrm{10}\right)}{\mathrm{sin}\:\mathrm{20}} \…