Question Number 203471 by DEGWE last updated on 19/Jan/24 Answered by MathematicalUser2357 last updated on 20/Jan/24 $$=\frac{\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{x}^{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}} }{\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}\right)} \\ $$$$=\frac{\underset{{x}\rightarrow\mathrm{0}^{+} }…
Question Number 203434 by sonukgindia last updated on 19/Jan/24 Answered by witcher3 last updated on 20/Jan/24 $$=\int\frac{−\mathrm{2sin}\left(\mathrm{2x}\right)\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)}\mathrm{dx} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{3}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)−\frac{\mathrm{1}}{\mathrm{6}}\int\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)\mathrm{dx} \\ $$$$\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{3x}\right)\right)=\mathrm{ln}\left(\mathrm{1}+\mathrm{2cos}\left(\mathrm{x}\right)\left(\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{1}\right)−\mathrm{4}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\mathrm{cos}\left(\mathrm{x}\right)\right) \\ $$$$=\mathrm{ln}\left(\mathrm{1}+\mathrm{8cos}^{\mathrm{3}}…
Question Number 203460 by MrGHK last updated on 19/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203457 by DEGWE last updated on 19/Jan/24 Answered by witcher3 last updated on 19/Jan/24 $$\mathrm{tanh}\left(\mathrm{x}\right)=\frac{\mathrm{e}^{\mathrm{2x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2x}} +\mathrm{1}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\frac{\mathrm{1}+\mathrm{3}^{\mathrm{x}} +\mathrm{x}^{\mathrm{3}} +\mathrm{sh}\left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{3}^{\mathrm{x}} −\mathrm{x}^{\mathrm{3}} −\mathrm{sh}\left(\mathrm{x}\right)}−\mathrm{1}}{\frac{\mathrm{1}+\mathrm{3}^{\mathrm{x}}…
Question Number 203389 by sonukgindia last updated on 18/Jan/24 Answered by mr W last updated on 18/Jan/24 Commented by mr W last updated on 18/Jan/24…
Question Number 203414 by professorleiciano last updated on 18/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203396 by Harkoriola last updated on 19/Jan/24 $$\mathrm{State}\:\mathrm{the}\:\mathrm{completness}\:\mathrm{axioms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{real}\:\mathrm{number} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203392 by professorleiciano last updated on 18/Jan/24 Commented by a.lgnaoui last updated on 19/Jan/24 Commented by a.lgnaoui last updated on 20/Jan/24 $$\bigtriangleup\boldsymbol{\mathrm{OAB}}\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} =\mathrm{41}−\mathrm{40cos}\:\boldsymbol{\mathrm{x}}…
Question Number 203393 by professorleiciano last updated on 18/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203394 by professorleiciano last updated on 18/Jan/24 Answered by som(math1967) last updated on 18/Jan/24 $$\frac{\bigtriangleup{BPQ}}{\bigtriangleup{ABC}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{BQ}}{{BC}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{BQ}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Rightarrow{BQ}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\:{cm} \\ $$ Terms of…