trig q (1 Viewer)

[shkspeare]

hi guys some1 help me out here

1) (tan2x-tanx) / (tan2x+cotx) = tan^2(x)
2) (1-tanxtan2x) / (1+tanxtan2x) = 4cos^2x-3

how to prove lhs = rhs here? thx !!

 

[CM_Tutor]

Result: (tan 2x - tan x) / (tan2x + cot x) = tan²x

Proof: LHS = (tan 2x - tan x) / (tan2x + cot x)
= {[2tan x / (1 - tan²x)] - tan x} / {[2tan x / (1 - tan²x)] + (1 / tan x)}

This is a quadruple decker fraction - yuk - so multiply top and bottom by tan x * (1 - tan²x)

= [2tan²x - tan²x(1 - tan²x)] / [2tan²x + (1 - tan²x)]
= tan²x(2 - 1 + tan²x) / (tan²x + 1)
= tan²x
= RHS

For the second result, the method should be the same - expand tan 2x as 2tan x / (1 - tan²x), multiply to get rid of quadruple decker fractions, and clean up.

 

[shkspeare]

wOWWW thx!!!