Question Number 185732 by Rupesh123 last updated on 26/Jan/23 Answered by witcher3 last updated on 26/Jan/23 $$\mathrm{tg}\left(\mathrm{a}\right)−\mathrm{tg}\left(\mathrm{b}\right)=\frac{\mathrm{sin}\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}\right)\mathrm{cos}\left(\mathrm{b}\right)} \\ $$$$\mathrm{tg}\left(\mathrm{50}\right)−\mathrm{tg}\left(\mathrm{40}\right)=\frac{\mathrm{sin}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{cos}\left(\mathrm{50}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{cos}\left(\mathrm{50}\right)}=\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{cos}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{40}\right)}=\mathrm{2}\frac{\mathrm{cos}\left(\mathrm{10}\right)}{\mathrm{2cos}\left(\mathrm{40}\right)\mathrm{sin}\left(\mathrm{40}\right)} \\ $$$$=\frac{\mathrm{2cos}\left(\mathrm{10}\right)}{\mathrm{sin}\left(\mathrm{80}\right)}=\mathrm{2} \\ $$…
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Question Number 54626 by Aditya789 last updated on 08/Feb/19 $$\frac{\mathrm{1}+\mathrm{sinx}+\mathrm{cox}}{\mathrm{1}+\mathrm{sinx}−\mathrm{cosx}}=\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{1}+\mathrm{cosx}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19 $${LHS} \\ $$$$\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}} \\…
Question Number 120156 by mnjuly1970 last updated on 29/Oct/20 $$\:\:\:\:\:\:\:\:…\spadesuit\:{nice}\:\:{calculus}\spadesuit… \\ $$$$\:\:{if}\:\:{cos}^{−\mathrm{1}} \left({x}\right)+{cos}^{−\mathrm{1}} \left({y}\right)+{cos}^{−\mathrm{1}} \left({z}\right)=\pi\checkmark \\ $$$$ \\ $$$$\:\:\:\:\:\:{show}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xyz}=\mathrm{1}\:\:\checkmark \\…
Question Number 185647 by Mingma last updated on 24/Jan/23 Answered by mahdipoor last updated on 24/Jan/23 $${secx}+{tan}^{\mathrm{3}} {x}.{cscx}=\frac{\mathrm{1}}{{cosx}}+\frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}=\frac{{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{3}} {x}}= \\ $$$$\frac{\mathrm{1}}{{cos}^{\mathrm{3}}…
Question Number 54563 by 951172235v last updated on 06/Feb/19 $$\mathrm{If}\:\frac{\mathrm{sin}\:^{\mathrm{3}} \alpha}{\mathrm{sin}\:\beta}\:+\:\frac{\mathrm{cos}\:^{\mathrm{3}} \alpha}{\mathrm{cos}\:\beta}\:=\:\mathrm{1}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{2sin}\:\left(\alpha+\beta\right)=\mathrm{0} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 06/Feb/19 $$\frac{{sin}^{\mathrm{3}} \alpha}{{sin}\beta}+\frac{{cos}^{\mathrm{3}}…
Question Number 54537 by hsbebeb last updated on 05/Feb/19 $${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{33}\:\:\: \\ $$$$\left({please}\:{solve}\:{this}\:{and}\:{i}\:{want}\:{to}\:{know}\:\right. \\ $$$$\left.{that}\:{which}\:{standard}\:{that}\:{question}\:{belongs}?\right) \\ $$$${help}\:{needed} \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 54536 by 951172235v last updated on 05/Feb/19 $$\mathrm{If}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{are}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\mathrm{Acot}\:\mathrm{B}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\mathrm{Bcot}\:\mathrm{C}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\mathrm{Ccot}\:\mathrm{A}\right) \\ $$$$=\:\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{1}+\frac{\mathrm{8cos}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2B}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2C}}\right\} \\ $$ Answered by…
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Question Number 120029 by Ar Brandon last updated on 28/Oct/20 $$\mathrm{Montrer}\:\mathrm{que}\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{cos}\left(\mathrm{sinx}\right)>\mathrm{sin}\left(\mathrm{cosx}\right) \\ $$ Commented by soumyasaha last updated on 29/Oct/20 $$\:\:\:\:\pi/\mathrm{2}\:\approx\:\mathrm{1}.\mathrm{57}\:\:\:\mathrm{and}\:\sqrt{\mathrm{2}}\:\approx\:\mathrm{1}.\mathrm{41} \\ $$$$\:\:\therefore\:\sqrt{\mathrm{2}}\:<\:\pi/\mathrm{2}…