Question

sin theta =2x/1+x^2 then tan theta​

Answer 1

Answer:

(From attachment diagram)

Solution :

sin theta = 2x / 1+x^2 = Adjacent/Hypotenuse

By phythagoras theorem

Base = √Hypotenuse ^2 - Adjacent^2

AB = √ (1+x^2) - (4x^2)

Ab = √1+x^4+2x^2-4x^2

AB = √1+x^4 -2x^2

AB= √(1-x^2)^2

AB= 1-x^2

Tan theta = BC/AB

= 1-x^2/2x

Answer: Tan theta = 1-x^2/2x

Happy independence Day

Answer 2

Answer:

$\mathrm{tan\theta = \dfrac{2x}{1 - x^2} }$

Step-by-step explanation:

Given :

$\mathrm {sin\theta = \dfrac{2x}{1 + x^2}}$

To find :

$\text{The value of } \mathrm{tan\theta}$

Solution :

Let us substitute x = tanФ

thus, Ф = tan⁻¹(x)

$\longmapsto \mathrm {sin\theta = \dfrac{2x}{1 + x^2}}$

So, on substituting, we get,

$\implies \mathrm {sin\theta = \dfrac{2tan\phi}{1 + tan^2\phi}}$

We know that,

$\boxed{\mathrm{ \dfrac{2tanx}{1 + tan^2x} = sin2x}}$

So,

$\implies \mathrm {sin\theta = sin2\phi}$

$\implies \theta = 2\phi$

As, Ф = tan⁻¹(x)

$\implies \mathrm{\theta = 2tan^{-1}x}$

$\implies \mathrm {tan^{-1}x = \dfrac{\theta}{2} }$

$\implies \mathrm {x = tan{\dfrac{\theta}{2} }}$

We know that,

$\boxed{\mathrm{tanx = \dfrac{2tan\dfrac{\theta}{2}}{1 - tan^2\dfrac{\theta}{2} } }}$

Substituting x in place of tan(θ/2), we get,

$\implies \underline{\boxed{\mathrm{tan\theta = \dfrac{2x}{1-x^2} }}}$

Hope it helps!!