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Question Number 118004 by TANMAY PANACEA last updated on 14/Oct/20

find value of tan46^0 using calculus

$${find}\:\:{value}\:{of}\:{tan}\mathrm{46}^{\mathrm{0}} \:{using}\:{calculus} \\ $$

Answered by mr W last updated on 14/Oct/20

46°=45°+1°=(π/4)+(π/(180)) tan 46°=((1+tan (π/(180)))/(1−tan (π/(180)))) ≈((1+(π/(180)))/(1−(π/(180))))=((180+π)/(180−π))=1.0355 or y=tan x ((Δy)/(Δx))≈(dy/dx)=(1/(cos^2 x)) Δy=tan 46°−tan 45°≈((Δx)/(cos^2 45°)) tan 46°≈tan 45°+(1/(cos^2 45°))×(π/(180)) ≈1+2×(π/(180))=1+(π/(90))=1.0349

$$\mathrm{46}°=\mathrm{45}°+\mathrm{1}°=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{180}} \\ $$$$\mathrm{tan}\:\mathrm{46}°=\frac{\mathrm{1}+\mathrm{tan}\:\frac{\pi}{\mathrm{180}}}{\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{180}}} \\ $$$$\approx\frac{\mathrm{1}+\frac{\pi}{\mathrm{180}}}{\mathrm{1}−\frac{\pi}{\mathrm{180}}}=\frac{\mathrm{180}+\pi}{\mathrm{180}−\pi}=\mathrm{1}.\mathrm{0355} \\ $$$${or} \\ $$$${y}=\mathrm{tan}\:{x} \\ $$$$\frac{\Delta{y}}{\Delta{x}}\approx\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}} \\ $$$$\Delta{y}=\mathrm{tan}\:\mathrm{46}°−\mathrm{tan}\:\mathrm{45}°\approx\frac{\Delta{x}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{45}°} \\ $$$$\mathrm{tan}\:\mathrm{46}°\approx\mathrm{tan}\:\mathrm{45}°+\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{45}°}×\frac{\pi}{\mathrm{180}} \\ $$$$\approx\mathrm{1}+\mathrm{2}×\frac{\pi}{\mathrm{180}}=\mathrm{1}+\frac{\pi}{\mathrm{90}}=\mathrm{1}.\mathrm{0349} \\ $$

Commented by TANMAY PANACEA last updated on 14/Oct/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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

$${sir}\:{encourage}\:{people}\:{to}\:{post}\:{logic}\:{based}\: \\ $$$${and}\:{interesting}\:{question}. \\ $$$${people}\:{are}\:{posting}\:{M}.{Sc}\:{level}\:{text}\:{book}\:{question} \\ $$

Answered by TANMAY PANACEA last updated on 14/Oct/20

y=tanx →(dy/dx)=sec^2 x (dy/dx)≈((△y)/(△x)) here x=45^o =(π/4), x+△x=46^o =(π/4)+(π/(180))=((46π)/(180)) now when x=(π/4) y=tanx=1 when x+△x=((46π)/(180)) y+△y=? here y=tanx=tan45^o =1 △y=(dy/dx)×△x=sec^2 x×△x △y=sec^2 ((π/4))×(π/(180))=(π/(90)) y+△y=1+(π/(90))

$${y}={tanx}\:\:\rightarrow\frac{{dy}}{{dx}}={sec}^{\mathrm{2}} {x} \\ $$$$\frac{{dy}}{{dx}}\approx\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$${here}\:{x}=\mathrm{45}^{{o}} =\frac{\pi}{\mathrm{4}},\:\:{x}+\bigtriangleup{x}=\mathrm{46}^{{o}} =\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{180}}=\frac{\mathrm{46}\pi}{\mathrm{180}} \\ $$$${now}\:{when}\:{x}=\frac{\pi}{\mathrm{4}}\:\:{y}={tanx}=\mathrm{1} \\ $$$${when}\:{x}+\bigtriangleup{x}=\frac{\mathrm{46}\pi}{\mathrm{180}}\:\:\:{y}+\bigtriangleup{y}=? \\ $$$${here}\:{y}={tanx}={tan}\mathrm{45}^{{o}} =\mathrm{1} \\ $$$$\bigtriangleup{y}=\frac{{dy}}{{dx}}×\bigtriangleup{x}={sec}^{\mathrm{2}} {x}×\bigtriangleup{x} \\ $$$$\bigtriangleup{y}={sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)×\frac{\pi}{\mathrm{180}}=\frac{\pi}{\mathrm{90}} \\ $$$${y}+\bigtriangleup{y}=\mathrm{1}+\frac{\pi}{\mathrm{90}} \\ $$$$ \\ $$

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