Question Number 24971 by Rishabh#1 last updated on 30/Nov/17 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:\mathrm{3}{x}}+\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{9}{x}}+\frac{\mathrm{cos}\:\mathrm{9}{x}}{\mathrm{cos}\:\mathrm{27}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:\mathrm{27}{x}−\mathrm{tan}\:{x}\right). \\ $$ Answered by math solver last updated on 09/Feb/18 $$\mathrm{assuming}\:\mathrm{3rd}\:\mathrm{term}\:\mathrm{to}\:\mathrm{be}\: \\ $$$$\:\:\Rightarrow\:\frac{\mathrm{sin9}{x}}{{cos}\mathrm{27}{x}}.…
Question Number 24972 by Rishabh#1 last updated on 30/Nov/17 $$\mathrm{If}\:\alpha,\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{conected}\:\mathrm{by}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\mathrm{2tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:+ \\ $$$$\:\:\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \gamma\:\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{which}\:\mathrm{of}\:\mathrm{these}\:\mathrm{are}\:\mathrm{correct}\left(\mathrm{multi}\:\mathrm{correct}\right)…
Question Number 90507 by student work last updated on 24/Apr/20 $$ \\ $$$$\:\mathrm{if}\:\:\mathrm{2}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{2}^{\mathrm{cos}\:\mathrm{x}} =\mathrm{2}^{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}=? \\ $$ Commented by student work last updated on…
Question Number 24959 by adityapratap2585@gmail.com last updated on 29/Nov/17 $$\mathrm{In}\:\mathrm{a}\:\bigtriangleup\mathrm{ABC},\:\angle\mathrm{B}=\pi/\mathrm{6},\:\angle\mathrm{C}=\pi/\mathrm{4}\:\mathrm{and} \\ $$$$\mathrm{D}\:\mathrm{divides}\:\mathrm{BC}\:\mathrm{internally}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\: \\ $$$$\mathrm{1}\::\:\mathrm{3}\:\mathrm{then}\:,\:\:\frac{\mathrm{sin}\:\angle\mathrm{BAD}}{\mathrm{sin}\:\angle\mathrm{CAD}}\:\:\:\mathrm{equal}\:\mathrm{to\_\_} \\ $$ Answered by myintkhaing last updated on 30/Nov/17 $$\frac{\sqrt{\mathrm{2}}}{\mathrm{6}} \\…
Question Number 156024 by alcohol last updated on 07/Oct/21 Answered by mr W last updated on 07/Oct/21 $$\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)} \\…