# Show that sin 2x=2tan x/(1+tan^2x)?

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L:H:S = sin2x

= 2sinx.cosx ÷ 1

Devide the numerator and denominator by cos²x

= (2sinx.cosx/cos²x) ÷ (1/cos²x)

= 2sinx/cosx ÷ sec²x

= 2tanx ÷ (1+tan²x)

= R:H:S

We'll manage the LHS.

We'll write the formula that writes the sine of double angle in terms of sine and cosine of the angle.

sin 2x = 2 sin x*cos x

Now, we'll multiply and divide by cos x the right side, to create the tangent function:

sin 2x = 2 sin x*cos x*cos x/cos x

sin 2x = 2 tan x*`cos^(2)` x

But, from Pythagorean identity, we'll have:

1 + `tan^(2)` x = 1/`cos^(2)` x => `cos^(2)` x = 1/(1+`tan^(2)` x)

sin 2x = 2 tan x/(1 + `tan^(2)` x)

**Therefore, the identity sin 2x = 2 tan x/(1 + `tan^(2)` x) is verified.**