find-value-of-tan46-0-using-calculus-

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

f i n d v a l u e o f t a n 46^{0} u s i n g c a l c u l u s

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

46 ° = 45 ° + 1 ° = \frac{π}{4} + \frac{π}{180}

\tan 46 ° = \frac{1 + \tan \frac{π}{180}}{1 - \tan \frac{π}{180}}

\approx \frac{1 + \frac{π}{180}}{1 - \frac{π}{180}} = \frac{180 + π}{180 - π} = 1.0355

o r

y = \tan x

\frac{Δ y}{Δ x} \approx \frac{d y}{d x} = \frac{1}{\cos^{2} x}

Δ y = \tan 46 ° - \tan 45 ° \approx \frac{Δ x}{\cos^{2} 45 °}

\tan 46 ° \approx \tan 45 ° + \frac{1}{\cos^{2} 45 °} \times \frac{π}{180}

\approx 1 + 2 \times \frac{π}{180} = 1 + \frac{π}{90} = 1.0349

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

t h a n k y o u s i r

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

s i r e n c o u r a g e p e o p l e t o p o s t l o g i c b a s e d

a n d i n t e r e s t i n g q u e s t i o n .

p e o p l e a r e p o s t i n g M . S c l e v e l t e x t b o o k q u e s t i o n

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

y = t a n x \to \frac{d y}{d x} = {s e c}^{2} x

\frac{d y}{d x} \approx \frac{△ y}{△ x}

h e r e x = 45^{o} = \frac{π}{4}, x + △ x = 46^{o} = \frac{π}{4} + \frac{π}{180} = \frac{46 π}{180}

n o w w h e n x = \frac{π}{4} y = t a n x = 1

w h e n x + △ x = \frac{46 π}{180} y + △ y = ?

h e r e y = t a n x = t a n 45^{o} = 1

△ y = \frac{d y}{d x} \times △ x = {s e c}^{2} x \times △ x

△ y = {s e c}^{2} (\frac{π}{4}) \times \frac{π}{180} = \frac{π}{90}

y + △ y = 1 + \frac{π}{90}