Question

PT
tan (π/4 + θ) – tan (π/4 –θ) = 2 tan ^2 θ

Answer 1

what's PT?

Answer 2

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

Answer 3

PT stands for Prove That

Answer 4

msingh , apply that formula for tan (π/4 + θ)

Answer 5

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

Answer 6

apply this formula for tan (π/4 - θ)

Answer 7

Simplify it~

Answer 8

let me know if u have any doubts further?

Answer 9

i m getting
4tan(theta) /(1-tan^theta)

Answer 10

@SandeepReddy

Answer 11

Yes msingh

Answer 12

After solving
i m getting
4tan(theta) /(1-tan^theta)

Answer 13

can anybody tell after this

Answer 14

according to wolframampha the identity is false
http://www.wolframalpha.com/input/?i=tan+%28%CF%80%2F4+%2B+%CE%B8%29+%E2%80%93+tan+%28%CF%80%2F4+%E2%80%93%CE%B8%29+%3D+2+tan+%5E2+%CE%B8

Answer 15

k
@experimentX
thank u

Answer 16

also can be seen through here
http://www.wolframalpha.com/input/?i=Plot+tan+%28%CF%80%2F4+%2B+%CE%B8%29+%E2%80%93+tan+%28%CF%80%2F4+%E2%80%93%CE%B8%29%2C+2+tan+%5E2+%CE%B8

Answer 17

yw

Answer 18

@experimentX what do u mean by "the identity is false"?

Answer 19

the left hand side is not equal to the right hand side.

Answer 20

Yeah @msingh must be writing it \[\tan (π/4 + θ) – \tan (π/4 –θ) = 2 \tan 2 θ\]

Answer 21

yup

Answer 22

my mistake

Answer 23

So u clear about it msingh?

Answer 24

so how to solve next

Answer 25

from 4tan(theta) /(1-tan^theta) ?

Answer 26

yes

Answer 27

from 4tan(theta) /(1-tan^2theta) ?

Answer 28

\[\frac{ 4\tan{\theta} }{ 1-\tan^2{\theta} }=2*\frac{ 2\tan{\theta} }{ 1-\tan^2{\theta} }\]
and we have a relation
\[\tan{2\theta} = \tan({\theta+\theta}) =\frac{ \tan{\theta}+\tan{\theta} }{ 1-\tan{\theta}*\tan{\theta} }=\frac{ 2\tan{\theta} }{ 1-\tan^2{\theta} }\]
therefore you have
\[\frac{ 4\tan{\theta} }{ 1-\tan^2{\theta} } = 2\tan{2\theta}\]

Answer 29

@SandeepReddy
thank yu
so much

Answer 30

My pleasure! ( and bit pain while writing equations though! :p )