1 + tanxtan(2x)=sec(2x)?

Question

please solve this identity and show work
thanks =]

Answer 1

1+tanxtan2x=1+tanx 2tanx/1-tan^2x
= 1-tan^2x+2tan^2x/1-tan^2x
= 1+tan^2x/1-tan^2x
= cos^2x +sin^2x/cos^2x-sin^2x
=1/cos2x
=sec2x

Answer 2

Prove the identity 1 + tanxtan(2x) = sec(2x).

Let's start with the left hand side.

1 + tanxtan(2x)
= 1 + tanx{2tanx/(1 - tan²x)}
= 1 + 2tan²x/(1 - tan²x)
= {(1 - tan²x) + 2tan²x}/(1 - tan²x)
= (1 + tan²x)/(1 - tan²x)
= sec²x/(1 - tan²x)
= 1/(cos²x - sin²x)
= 1/cos(2x)
= sec(2x) = Right Hand Side

Answer 3

1 + tan x tan 2x = 1 + (2tan^2 x)/(1-tan^2 x)
= [1-tan^2 x + 2tan^2 x]/(1-tan^2 x)
= (1+ tan^2 x) / (1-tan^2 x)
Multiplying both the numerator and denominator by cos^2 x gives,
= [cos^2 x + sin^2 x] / [cos^2 x - sin^2 x]
= 1/(cos^2 x - sin^2 x)
= 1/cos 2x = sec 2x

as required.

Answer 4

1 + tanx tan (2x) = 1 + tanx sin(2x) / cos(2x)

= 1 + ( sin x / cos x ) 2 sin x cos x / (2 cos² x - 1)

= 1 + 2sin² x / (2cos² x - 1)

= ((2cos² x - 1) + 2 sin² x)) / (2cos²x - 1)

= (2(sin²x + cos²x) - 1) / (2cos²x - 1)

= (2 - 1 ) / (2cos² - 1 ) = 1/(2cos²x - 1)

= 1/cos 2x = sec 2x

Answer 5

LHS=1+(sinx*sin2x)/(cosx*cos2x)
=(cosxcos2x+sinxsin2x)/(cosx*cos2x)
=cosx(cos2x+2*sinx*sinx)/(cosx*cos2x)
=(cosx*cosx-sinx*sinx+2*sinx*sinx)/cos2x
=(cosx*cosx+sinx*sinx)/cos2x
=1/cos2x=sec(2x)