Question Number 156024 by alcohol last updated on 07/Oct/21 Answered by mr W last updated on 07/Oct/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{f}\left(\mathrm{3}{x}\right)}}{{x}} \\ $$$$=\mathrm{3}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{f}\left(\mathrm{3}{x}\right)}}{\mathrm{3}{x}} \\ $$$$=\mathrm{3}\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{f}\left({t}\right)}}{{t}}…
Question Number 24948 by adityapratap2585@gmail.com last updated on 29/Nov/17 $$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{the}\:\mathrm{moon}'\mathrm{s}\:\mathrm{diameter}\: \\ $$$$\mathrm{subtends}\:\mathrm{and}\:\mathrm{angle}\:\left(\mathrm{1}/\mathrm{2}\right)°\:\mathrm{at}\:\mathrm{the}\:\mathrm{eye}\: \\ $$$$\mathrm{of}\:\mathrm{an}\:\mathrm{observer},\:\mathrm{find}\:\mathrm{how}\:\mathrm{far}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{eye}\:\mathrm{of}\:\mathrm{a}\:\mathrm{coin}\:\mathrm{of}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{diameter}\:\mathrm{must}\: \\ $$$$\mathrm{be}\:\mathrm{held}\:\mathrm{so}\:\mathrm{as}\:\mathrm{just}\:\mathrm{to}\:\mathrm{hide}\:\mathrm{moon}\:? \\ $$ Commented by adityapratap2585@gmail.com last updated…
Question Number 24933 by adityapratap2585@gmail.com last updated on 29/Nov/17 $$\mathrm{if}\:\mathrm{6sin}^{\mathrm{4}} \theta+\mathrm{3cos}^{\mathrm{4}} \theta=\mathrm{2}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\left(\mathrm{7cosec}^{\mathrm{6}} \theta+\mathrm{8sec}^{\mathrm{6}} \theta\right)^{\mathrm{1}/\mathrm{3}} \\ $$ Commented by prakash jain last updated on…
Question Number 155989 by cortano last updated on 07/Oct/21 $$\:\:\:\:\mathrm{5sin}\:^{\mathrm{2}} \mathrm{2x}\:+\:\mathrm{8cos}\:^{\mathrm{3}} \mathrm{x}\:=\:\mathrm{8cos}\:\mathrm{x} \\ $$$$\:\:\:\:\frac{\mathrm{3}\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$ Commented by john_santu last updated on 07/Oct/21 $$\:{x}=\frac{\mathrm{3}\pi}{\mathrm{2}},\:\mathrm{2}\pi−\mathrm{arccos}\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{2}\pi \\…
Question Number 24912 by tawa tawa last updated on 28/Nov/17 Answered by mrW1 last updated on 29/Nov/17 $$\left({i}\right) \\ $$$${L}=\frac{\mathrm{2}×\mathrm{36}}{\mathrm{180}}×\pi{R}=\frac{\mathrm{2}×\mathrm{36}}{\mathrm{180}}×\frac{\mathrm{22}}{\mathrm{7}}×\mathrm{6400}=\mathrm{8050}\:{km} \\ $$$$\left({ii}\right) \\ $$$${t}=\frac{{L}}{{v}}=\frac{\mathrm{8050}}{\mathrm{800}}=\mathrm{10}\:{h} \\…
Question Number 155898 by Tawa11 last updated on 05/Oct/21 $$\mathrm{Show}\:\mathrm{that}\:\:\:\:\:\mathrm{tan}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:\:+\:\:\:\mathrm{sin}^{−\:\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:\:\:=\:\:\:\frac{\pi}{\mathrm{4}} \\ $$ Answered by immortel last updated on 05/Oct/21 $${L}={tan}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)=\frac{{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)+{tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−{tan}\left({arctan}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right){tan}\left({arcsin}\frac{\mathrm{1}}{\mathrm{3}}\right)} \\…
Question Number 155810 by zainaltanjung last updated on 05/Oct/21 $$\mathrm{Verify}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{in}\:\mathrm{Excercise}\:\mathrm{below} \\ $$$$\left.\mathrm{1}\right).\:\mathrm{cos}\:\theta\mathrm{sec}\:\theta=\mathrm{1} \\ $$$$\left.\mathrm{2}\right).\:\left(\mathrm{1}+\mathrm{cos}\:\beta\right)\left(\mathrm{1}−\mathrm{cos}\:\beta\right)=\mathrm{sin}\:^{\mathrm{2}} \beta \\ $$$$\left.\mathrm{3}\right).\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)=\mathrm{sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\left.\mathrm{4}\right).\:\frac{\mathrm{sin}\:\mathrm{t}}{\mathrm{cosec}\:\mathrm{t}}+\frac{\mathrm{cos}\:\mathrm{t}}{\mathrm{sec}\:\mathrm{t}}=\mathrm{1} \\ $$$$\left.\mathrm{5}\right).\:\frac{\mathrm{cosec}\:^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}}…
Question Number 24554 by tawa tawa last updated on 20/Nov/17 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{2q}}\right)\:+\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{p}}{\mathrm{p}\:+\:\mathrm{q}}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$ Commented by mrW1 last updated on 21/Nov/17 $${that}'{s}\:{not}\:{true}! \\ $$$${let}'{s}\:{say}\:{p}={q}=\mathrm{1}…
Question Number 89958 by 20000193 last updated on 20/Apr/20 $${prove}\:{that}\:\left(\mathrm{1}+\mathrm{sin}\:{x}/\mathrm{1}+\mathrm{cos}\:\mathrm{3}\left(\mathrm{1sin}\:{x}/\mathrm{1}+\mathrm{cosec}\:{x}\right)={tanx}\right. \\ $$ Commented by john santu last updated on 20/Apr/20 $$\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{3}\left(\mathrm{1sin}\:{x}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosec}\:{x}\:}\right)}\:=\:\mathrm{tan}\:{x}? \\ $$ Terms of…
Question Number 89951 by john santu last updated on 20/Apr/20 $$\mathrm{log}\:_{\mathrm{2}} \:\left(\mathrm{sin}\:\left({x}+\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right)\:+\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{12}}\right)\right)=−\mathrm{1} \\ $$ Commented by jagoll last updated on 20/Apr/20 $$\Rightarrow\mathrm{sin}\:\left(\mathrm{x}+\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\:\mathrm{sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{12}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{12}}\right)−\mathrm{cos}\:\left(\mathrm{2x}+\frac{\mathrm{6}\pi}{\mathrm{12}}\right)\:=\:\mathrm{1}…