Question

tan9A -tan7A -tan2A=tan9A.tan7A.tan2A​

Answer 1

For this equation we will use the formula:

i) tan 2a= tan(a + a) and,

ii)tan (a + b)= (tan a+ tan b)/(1-tan a tan b)

From (i), we can say

tan 13A= tan (9A+4A)

but, tan(9A+4A)=(tan 9A + tan 4A)/ (1-tan 9Atan 4A)

Therefore tan13A=(tan 9A+tan4A)/(1-tan9Atan4A)

=tan13A-tan13Atan9Atan4A=tan 9A+tan 4A

therefore,

tan13A-tan9A-tan4A=tan13Atan9Atan4A