Question

answer the question​

Answer 1

Answer:

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

Step-by-step explanation:

Given :

$\large cos \ \theta=\dfrac{2x}{1+x^2}$

We have to find tan θ

We know that

$\large cos \ \theta=\dfrac{Base}{Hypotenuse} \\\\\\\large We \ have \ Pythagoras' \ theorem\\\\\\\large (Hypotenuse)^2=(Base)^2+(Perpendicular)^2\\\\\\\large Putting \ value \ here \ we \ get\\\\\\\large (1+x^2)^2=(2x)^2+(Perpendicular)^2\\\\\\\large (1+x^4+2x^2)=4x^2+(Perpendicular)^2\\\\\\\large (Perpendicular)^2=1+x^4+2x^2-4x^2\\\\\\\large (Perpendicular)^2=1+x^4-2x^2\\\\\\\large (Perpendicular)^2=(1-x^2)^2$

$\large (Perpendicular)=(1-x^2)^\\\\\\\large For \ tan \ \theta =\dfrac{Perpendicular}{Base} \\\\\\\large Put \ the \ value \ here\\\\\\\large tan \ \theta =\dfrac{ 1-x^2}{2x}$

Thus we get answer.