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Question-140192

Question Number 140192 by meetbhavsar25 last updated on 05/May/21 Answered by mr W last updated on 05/May/21 $${say}\:\mathrm{tan}^{−\mathrm{1}} {x}={t} \\ $$$${x}=\mathrm{tan}\:{t} \\ $$$$\mathrm{cos}\:{t}=\mathrm{tan}\:{t}=\frac{\mathrm{sin}\:{t}}{\mathrm{cos}\:{t}} \\ $$$$\mathrm{cos}^{\mathrm{2}}…

please-integrate-f-x-0-1-1-z-log-z-2-2zcos-x-1-z-1-2-dz-

Question Number 140194 by mnjuly1970 last updated on 05/May/21 $$\:\: \\ $$$$\:\:\:{please}\:\:{integrate}:: \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{\mathrm{1}}{{z}}{log}\left(\frac{{z}^{\mathrm{2}} +\mathrm{2}{zcos}\left({x}\right)+\mathrm{1}}{\left({z}+\mathrm{1}\right)^{\mathrm{2}} }\right)\right\}{dz} \\ $$$$ \\ $$ Answered by Dwaipayan…

If-x-5-1-4-5-1-4-and-y-5-1-4-5-1-4-Show-that-5-x-2-y-2-2-144-

Question Number 9119 by tawakalitu last updated on 19/Nov/16 $$\mathrm{If}\:\:\mathrm{x}\:=\:\mathrm{5}^{\mathrm{1}/\mathrm{4}} \:+\:\mathrm{5}^{−\mathrm{1}/\mathrm{4}} \:\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\mathrm{5}^{\mathrm{1}/\mathrm{4}} \:−\:\mathrm{5}^{−\mathrm{1}/\mathrm{4}} \\ $$$$\mathrm{Show}\:\mathrm{that}\::\:\mathrm{5}^{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} \:} =\:\mathrm{144} \\ $$ Answered by mrW last…

Question-74655

Question Number 74655 by TawaTawa last updated on 28/Nov/19 $$. \\ $$ Commented by TawaTawa last updated on 28/Nov/19 The force F acting along an inclined plane is just sufficient to maintain a body on the plane, the angle of friction M being less than Y, the angle of plane. prove that the least force acting along the plane, sufficient to drag the body up the plane is : F sin( M + Y )/sin( M - Y) Answered by mr W last…

Solution-equation-sin2x-1-2-cosx-cos2x-

Question Number 140187 by mathsuji last updated on 05/May/21 $${Solution}\:{equation}: \\ $$$${sin}\mathrm{2}{x}=\mathrm{1}+\sqrt{\mathrm{2}}\:{cosx}+{cos}\mathrm{2}{x} \\ $$ Answered by Ankushkumarparcha last updated on 05/May/21 $${Solution}:\:\mathrm{sin}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2cos}^{\mathrm{2}} \left({x}\right)+\sqrt{\mathrm{2}}\mathrm{cos}\left({x}\right)\:\left(\because\:\mathrm{cos}\left(\mathrm{2}{x}\right)\:=\:\mathrm{2cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}\right) \\…