Category: Trigonometry

Question Number 112885 by malwan last updated on 10/Sep/20 Commented by malwan last updated on 10/Sep/20 $$P=?$$$$\left(a\right)(-5\pi /4,2^{1/2})$$$$\left(b\right)(-5\pi /4,2^{-1/2})$$$$\left({c}\right)\:\left(−\mathrm{7}\pi/\mathrm{4},\mathrm{2}^{\mathrm{1}/\mathrm{2}}…

Question Number 47150 by somil last updated on 05/Nov/18 Commented by maxmathsup by imad last updated on 05/Nov/18 $$weseethatcos\theta mustbe⩾0\Rightarrow \theta \in [-\frac{\pi }{2},\frac{\pi }{2}]\left[2\pi \right]changement=x=sin\theta give$$$$\mathrm{3}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow\mathrm{9}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{4}{x}^{\mathrm{4}} \:\Rightarrow\mathrm{4}{x}^{\mathrm{4}}…

Question Number 178181 by mathlove last updated on 13/Oct/22 $$\frac{cos\alpha cot\alpha -sin\alpha tan\alpha }{csc\alpha sec\alpha }=?$$ Commented by Adedayo2000 last updated on 13/Oct/22 $$Recallcot\alpha =\frac{1}{tan\alpha }=\frac{cos\alpha }{sin\alpha },csc\alpha =\frac{1}{sin\alpha },sec\alpha =\frac{1}{cos\alpha }$$$$=>\frac{\left(cos\alpha \right)\left(\frac{cos\alpha }{sin\alpha }\right)-\left(sin\alpha \right)\left(\frac{sin\alpha }{cos\alpha }\right)}{\left(\frac{1}{sin\alpha }\right)\left(\frac{1}{cos\alpha }\right)}$$$$=>\frac{\frac{{cos}^{\mathrm{2}} \alpha}{{sin}\alpha}−\frac{{sin}^{\mathrm{2}}…

Question Number 178182 by mathlove last updated on 13/Oct/22 $$\frac{1}{cos80}-\frac{\sqrt{3}}{sin80}=?$$ Commented by cortano1 last updated on 13/Oct/22 $$\frac{1}{\sin 10°}-\frac{\sqrt{3}}{\cos 10°}=\frac{\cos 10°-\sqrt{3}\sin 10}{\sin 10°\cos 10°}$$$$=\frac{4(\frac{1}{2}\cos 10°-\frac{\sqrt{3}}{2}\sin 10°)}{\sin 20°}$$$$\:=\:\frac{\mathrm{4}\left(\mathrm{sin}\:\mathrm{30}°\:\mathrm{cos}\:\mathrm{10}°−\mathrm{cos}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{10}°\right)}{\mathrm{sin}\:\mathrm{20}°} \…

Question Number 178091 by alcohol last updated on 12/Oct/22 $$remainderwhen{7}^{9^{9^{9^9}}}isdividedby10$$ Terms of Service Privacy Policy Contact: info@tinkutara.com