Question Number 230 by ssahoo last updated on 25/Jan/15 $$\mathrm{Let}\:{z}\:\mathrm{and}\:{w}\:\mathrm{be}\:\mathrm{two}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mid{z}\mid\leqslant\mathrm{1}\:,\:\mid{w}\mid\leqslant\mathrm{1}\:\mathrm{and}\:\mid{z}+{iw}\mid=\mid{z}−{i}\overline {{w}}\mid=\mathrm{2}, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{z}. \\ $$ Commented by 123456 last updated on 16/Dec/14 $$\mid{z}+{iw}\mid\leqslant\mid{z}\mid+\mid{w}\mid\leqslant\mathrm{2}…
Question Number 65745 by KanhAshish last updated on 03/Aug/19 Commented by KanhAshish last updated on 03/Aug/19 $$\mathrm{15}\:\mathrm{and}\:\mathrm{16}\:\mathrm{both}\:\mathrm{solution}?? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 65743 by aliesam last updated on 03/Aug/19 Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19 $$\:\: \\ $$$${On}\:{a} \\…
Question Number 65724 by bshahid010@gmail.com last updated on 02/Aug/19 Answered by mr W last updated on 03/Aug/19 $${S}=\mathrm{sin}\:{x}+\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\mathrm{sin}\:{x}+\mathrm{cos}\:{x}+\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}}+\frac{\mathrm{1}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}} \\ $$$$=\mathrm{sin}\:{x}+\mathrm{cos}\:{x}+\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}+\mathrm{1}}{\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}} \\ $$$${let}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}={t} \\…
Question Number 65723 by bshahid010@gmail.com last updated on 02/Aug/19 Answered by Tanmay chaudhury last updated on 03/Aug/19 $$\mathrm{7}{a}=\mathrm{2}\pi \\ $$$$\mathrm{4}{a}=\mathrm{2}\pi−\mathrm{3}{a} \\ $$$${cos}\left(\mathrm{4}{a}\right)={cos}\left(\mathrm{2}\pi−\mathrm{3}{a}\right) \\ $$$$\left(\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{a}−\mathrm{1}\right)={cos}\mathrm{3}\pi…
Question Number 65680 by bshahid010@gmail.com last updated on 01/Aug/19 Answered by mr W last updated on 02/Aug/19 Commented by mr W last updated on 02/Aug/19…
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Question Number 131218 by shaker last updated on 02/Feb/21 Answered by mathmax by abdo last updated on 02/Feb/21 $$\mathrm{I}\:=\int_{−\mathrm{2}} ^{\mathrm{2}} \:\frac{\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{4}^{\mathrm{x}} \:+\mathrm{1}}\:\Rightarrow\mathrm{I}\:=_{\mathrm{x}=−\mathrm{t}} \:\:\:\int_{−\mathrm{2}} ^{\mathrm{2}}…
Question Number 131214 by shaker last updated on 02/Feb/21 Answered by Ar Brandon last updated on 02/Feb/21 $$\mathrm{A}=\sqrt{\mathrm{3}\sqrt{\mathrm{5}\sqrt{\mathrm{3}\sqrt{\mathrm{5}…}}}}\:\Rightarrow\mathrm{A}=\sqrt{\mathrm{3}\sqrt{\mathrm{5A}}} \\ $$$$\Rightarrow\mathrm{A}^{\mathrm{4}} =\mathrm{9}×\mathrm{5A}\:\Rightarrow\mathrm{A}^{\mathrm{4}} −\mathrm{45A}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{A}\left(\mathrm{A}^{\mathrm{3}} −\mathrm{45}\right)=\mathrm{A}\left(\mathrm{A}^{\mathrm{3}}…
Question Number 131201 by shaker last updated on 02/Feb/21 Answered by liberty last updated on 11/Feb/21 $$\mathrm{L}=\int\:\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:\mathrm{2x}}\:\mathrm{dx}\:;\:\mathrm{by}\:\mathrm{change}\:\mathrm{of}\:\mathrm{variable}\: \\ $$$$\:\mathrm{let}\:\mathrm{u}\:=\:\mathrm{tan}\:\mathrm{x}\:\rightarrow\begin{cases}{\mathrm{dx}=\frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }}\\{\mathrm{cos}\:\mathrm{2x}=\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }}\end{cases} \\ $$$$\mathrm{L}=\int\:\left(\frac{\mathrm{u}^{\mathrm{2}}…