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Author: Tinku Tara

Question-211190

Question Number 211190 by mr W last updated on 30/Aug/24 Answered by A5T last updated on 31/Aug/24 $${Let}\:\angle{BCD}=\angle{CBD}=\theta\Rightarrow\angle{DBA}=\mathrm{90}°−\theta \\ $$$$\Rightarrow\angle{ADB}=\mathrm{90}°−\theta\Rightarrow\angle{DAB}=\mathrm{2}\theta \\ $$$$\angle{BCA}=\angle{CAB}=\mathrm{45}°\Rightarrow\angle{ACD}=\mathrm{45}°−\theta\Rightarrow?=\mathrm{45}°−\mathrm{2}\theta \\ $$$${Let}\:{CD}={BD}={x}\:{and}\:{AD}={AB}={BC}={y} \\…

sin-n-sin-1-n-cos-sin-n-1-d-

Question Number 211155 by Spillover last updated on 30/Aug/24 $$\int\frac{\left(\mathrm{sin}\:^{{n}} \left(\theta\right)−\mathrm{sin}\:\left(\theta\right)\right)^{\frac{\mathrm{1}}{{n}}} \mathrm{cos}\:\left(\theta\right)}{\mathrm{sin}\:^{{n}+\mathrm{1}} \left(\theta\right)}{d}\theta \\ $$$$ \\ $$ Answered by Frix last updated on 30/Aug/24 $$=\int\frac{{d}\theta}{\mathrm{sin}\:\theta}\:−\int\left(\mathrm{sin}\:\theta\right)^{\frac{\mathrm{1}−{n}−{n}^{\mathrm{2}}…

Question-211180

Question Number 211180 by Durganand last updated on 30/Aug/24 Answered by mr W last updated on 30/Aug/24 $$\left({ax}+{by}+{c}\right)\left(\frac{{x}}{{a}}+\frac{{y}}{{b}}+{d}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right){xy}+\left({ad}+\frac{{c}}{{a}}\right){x}+\left({bd}+\frac{{c}}{{b}}\right)+{cd} \\ $$$$={x}^{\mathrm{2}} +{y}^{\mathrm{2}}…