Question

why is 1-tanA divided by 1+tanA =(tanπ/4-A)​

Answer 1

Step-by-step explanation:

=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=

See The Attachment My Friend....

And Clear Your Doubt...

=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=×=

Answer 2

To Prove :

$\sf Tan( \frac{\pi}{4} - A) = \frac{1 -Tan(A) }{1 +Tan(A) }$

Proof :

$\Rightarrow \sf Tan( \frac{\pi}{4} - A) = \frac{Tan( \frac{\pi}{4}) \: - \: Tan(A )}{1 + Tan( \frac{\pi}{4})Tan(A) } \\ \\ \sf Because \: , \: \: Tan(x - y) = \frac{Tan(x) - Tan(y)}{1 + Tan(x)Tan(y)} \\ \\ \Rightarrow \sf Tan( \frac{\pi}{4} - A) = \frac{1 -Tan(A) }{1 +Tan(A) }$

Hence proved

Extra Information :

$\sf \star \: \: Tan(x - y) = \frac{Tan(x) - Tan(y)}{1 + Tan(x)Tan(y)} \\ \\ \star \: \: </p><p>\sf Tan(x + y) = \frac{Tan(x) +Tan(y)}{1 - Tan(x)Tan(y)}</p><p>\\ \\ \star \: \: </p><p> \sf Tan( \frac{\pi}{4} + A) = \frac{1 +Tan(A) }{1 -Tan(A) }$