Question Number 118004 by TANMAY PANACEA last updated on 14/Oct/20
$${find}\:\:{value}\:{of}\:{tan}\mathrm{46}^{\mathrm{0}} \:{using}\:{calculus} \\ $$
Answered by mr W last updated on 14/Oct/20
$$\mathrm{46}°=\mathrm{45}°+\mathrm{1}°=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{180}} \\ $$$$\mathrm{tan}\:\mathrm{46}°=\frac{\mathrm{1}+\mathrm{tan}\:\frac{\pi}{\mathrm{180}}}{\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{180}}} \\ $$$$\approx\frac{\mathrm{1}+\frac{\pi}{\mathrm{180}}}{\mathrm{1}−\frac{\pi}{\mathrm{180}}}=\frac{\mathrm{180}+\pi}{\mathrm{180}−\pi}=\mathrm{1}.\mathrm{0355} \\ $$$${or} \\ $$$${y}=\mathrm{tan}\:{x} \\ $$$$\frac{\Delta{y}}{\Delta{x}}\approx\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\Delta{y}=\mathrm{tan}\:\mathrm{46}°−\mathrm{tan}\:\mathrm{45}°\approx\frac{\Delta{x}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{45}°} \\ $$$$\mathrm{tan}\:\mathrm{46}°\approx\mathrm{tan}\:\mathrm{45}°+\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{45}°}×\frac{\pi}{\mathrm{180}} \\ $$$$\approx\mathrm{1}+\mathrm{2}×\frac{\pi}{\mathrm{180}}=\mathrm{1}+\frac{\pi}{\mathrm{90}}=\mathrm{1}.\mathrm{0349} \\ $$
Commented by TANMAY PANACEA last updated on 14/Oct/20
$${thank}\:{you}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 14/Oct/20
$${sir}\:{encourage}\:{people}\:{to}\:{post}\:{logic}\:{based}\: \\ $$$${and}\:{interesting}\:{question}. \\ $$$${people}\:{are}\:{posting}\:{M}.{Sc}\:{level}\:{text}\:{book}\:{question} \\ $$
Answered by TANMAY PANACEA last updated on 14/Oct/20
$${y}={tanx}\:\:\rightarrow\frac{{dy}}{{dx}}={sec}^{\mathrm{2}} {x} \\ $$$$\frac{{dy}}{{dx}}\approx\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$${here}\:{x}=\mathrm{45}^{{o}} =\frac{\pi}{\mathrm{4}},\:\:{x}+\bigtriangleup{x}=\mathrm{46}^{{o}} =\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{180}}=\frac{\mathrm{46}\pi}{\mathrm{180}} \\ $$$${now}\:{when}\:{x}=\frac{\pi}{\mathrm{4}}\:\:{y}={tanx}=\mathrm{1} \\ $$$${when}\:{x}+\bigtriangleup{x}=\frac{\mathrm{46}\pi}{\mathrm{180}}\:\:\:{y}+\bigtriangleup{y}=? \\ $$$${here}\:{y}={tanx}={tan}\mathrm{45}^{{o}} =\mathrm{1} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}×\bigtriangleup{x}={sec}^{\mathrm{2}} {x}×\bigtriangleup{x} \\ $$$$\bigtriangleup{y}={sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)×\frac{\pi}{\mathrm{180}}=\frac{\pi}{\mathrm{90}} \\ $$$${y}+\bigtriangleup{y}=\mathrm{1}+\frac{\pi}{\mathrm{90}} \\ $$$$ \\ $$