Question Number 86558 by 71285 last updated on 29/Mar/20 $$\mathrm{For}\:\mathrm{any}\:\mathrm{integer}\:{n},\underset{\:\mathrm{0}} {\overset{\pi} {\int}}{e}^{\mathrm{cos}^{\mathrm{2}} {x}} \mathrm{cos}^{\mathrm{3}} \left(\mathrm{2}{n}+\mathrm{1}\right){x}\:{dx}= \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86507 by ram roop sharma last updated on 29/Mar/20 $$\mathrm{If}\:{e}^{\mathrm{cos}\:{x}} −{e}^{−\mathrm{cos}\:{x}} =\:\mathrm{4},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{cos}\:{x}\:\mathrm{is} \\ $$ Commented by jagoll last updated on 29/Mar/20…
Question Number 86498 by ram roop sharma last updated on 29/Mar/20 $$\mathrm{If}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{x}+\mathrm{1}}}\left\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{4}{x}+\mathrm{1}}}{\mathrm{2}}\right)^{{n}} −\:\left(\frac{\mathrm{1}−\sqrt{\mathrm{4}{x}+\mathrm{1}}}{\mathrm{2}}\right)^{{n}} \right\} \\ $$$$\:\:\:\:=\:{a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+…+{a}_{\mathrm{5}} {x}^{\mathrm{5}} ,\:\mathrm{then}\:{n}= \\ $$ Terms of Service…
Question Number 86499 by ram roop sharma last updated on 29/Mar/20 $$\mathrm{The}\:\mathrm{digit}\:\mathrm{at}\:\mathrm{unit}'\mathrm{s}\:\mathrm{place}\:\mathrm{in}\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{17}^{\mathrm{1995}} +\:\mathrm{11}^{\mathrm{1995}} −\mathrm{7}^{\mathrm{1995}} \:\:\mathrm{is} \\ $$ Commented by Serlea last updated on…
Question Number 86496 by ram roop sharma last updated on 29/Mar/20 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\:\:\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}}} +\:\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{8}}} \right)^{\mathrm{1024}} \:\mathrm{is} \\ $$ Commented by Serlea last updated on…
Question Number 86490 by ram roop sharma last updated on 29/Mar/20 $$\mathrm{If}\:\:\mathrm{sin}\:\theta\:+\:\mathrm{cos}\:\theta=\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta\:\mathrm{then} \\ $$$$\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by john santu last updated on 29/Mar/20 $$\mathrm{sin}\:\theta\:=\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta−\mathrm{cos}\:\theta…
Question Number 20925 by dk3029943@gmail.com last updated on 08/Sep/17 $$\mathrm{In}\:\mathrm{a}\:\bigtriangleup{ABC},\:\mathrm{if}\:{a}=\mathrm{2},\:{b}=\mathrm{60}°\:\mathrm{and}\:{c}=\mathrm{75}°, \\ $$$$\mathrm{then}\:{b}\:= \\ $$ Answered by Joel577 last updated on 08/Sep/17 $$\angle{A}\:=\:\mathrm{180}°\:−\:\left(\mathrm{60}°\:+\:\mathrm{75}°\right)\:=\:\mathrm{45}° \\ $$$$ \\…
Question Number 20436 by pp75718276@gmail.com last updated on 26/Aug/17 $$\mathrm{If}\:\:{A}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{third}\:\mathrm{quadrant}\:\mathrm{and} \\ $$$$\mathrm{3}\:\mathrm{tan}\:{A}\:−\:\mathrm{4}\:=\:\mathrm{0},\:\mathrm{then}\: \\ $$$$\mathrm{5}\:\mathrm{sin}\:\mathrm{2}{A}\:+\:\mathrm{3}\:\mathrm{sin}\:{A}\:+\:\mathrm{4cos}\:{A}\:=\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20435 by pp75718276@gmail.com last updated on 26/Aug/17 $$\mathrm{If}\:\:{A}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{third}\:\mathrm{quadrant}\:\mathrm{and} \\ $$$$\mathrm{3}\:\mathrm{tan}\:{A}\:−\:\mathrm{4}\:=\:\mathrm{0},\:\mathrm{then}\: \\ $$$$\mathrm{5}\:\mathrm{sin}\:\mathrm{2}{A}\:+\:\mathrm{3}\:\mathrm{sin}\:{A}\:+\:\mathrm{4cos}\:{A}\:=\: \\ $$ Answered by Tinkutara last updated on 27/Aug/17 $$\mathrm{tan}\:{A}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:\mathrm{sin}\:{A}\:=\:\frac{−\mathrm{4}}{\mathrm{5}},\:\mathrm{cos}\:{A}\:=\:\frac{−\mathrm{3}}{\mathrm{5}} \\…
Question Number 20434 by pp75718276@gmail.com last updated on 26/Aug/17 $$\mathrm{If}\:\:\mathrm{cos}\:{x}=\mathrm{tan}\:{y},\:\mathrm{cos}\:{y}=\mathrm{tan}\:{z},\:\mathrm{cos}\:{z}=\mathrm{tan}\:{x}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{sin}\:{x}\:\:\mathrm{is} \\ $$ Answered by ajfour last updated on 27/Aug/17 $$\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {z}=\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {z}} \\…