Question

the value of tan(2tan^-1 (1/5)-pi/4) is?

Answer 1

Let\[\theta=2\tan^{-1} (1/5)-(\pi/4)\]Form a triangle with tan theta

Answer 2

how?

Answer 3

How what?!

Answer 4

wrong!

Answer 5

let x= \[\tan^{-1} \frac{1}{5}\]

Answer 6

so our expression becomes \[\tan(2x-\frac{\pi}{4})\]

Answer 7

now apply difference of angle formula

Answer 8

\[\tan(A-B) = \frac{\tan(A)-\tan(B) } {1+\tan(A)\tan(B) } \]

Answer 9

so we get

Answer 10

\[\frac{\tan(2x) -1}{1+\tan(2x)} \]

Answer 11

but ,\[\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)} \]

Answer 12

now, we go back to our substitution

x= tan^-1(1/5)

take tan of both sides , and we get tan(x) = 1/5

now we can draw up a general right angle triangle and mark an angle x , and fill in the lengths 1 and 5 for thre opposite and adjacent sides respectively

remember tan = opposite/adjacent

Answer 13

note the hypotenuse is sqrt(26)

Answer 14

now from that triangle we can find the value of sin(x) and cos(x) , the then you just plug them into the expression we had above

Answer 15

so \[\sin(x) = \frac{1}{\sqrt{26}}\] \[\cos(x) = \frac{5}{\sqrt{26}}\]

just by using the definitions

Answer 16

so , our whole expression was\[\tan(2x- \frac{\pi}{4}) = \frac{ \frac{ 2\sin(x)\cos(x) }{\cos^2(x)-\sin^2(x) } -1}{ 1+ \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}}\]

Answer 17

then sub in the values for sin(x) and cos(x) above, and simplify

Answer 18

you can do that

Answer 19

good. it worked out well. thanx man!