Question

answer the following​

Answer 1

Answer:

$\large \dfrac{1}{9}$

Step-by-step explanation:

Given :

$\large tan \ \theta=\dfrac{3}{4}$

$\large \text{We have to find \dfrac{1-cos \ \theta}{1+cos \ \theta} }$

We know that

$\large tan \ \theta=\dfrac{perpendicular}{base} \\\\\\\large We \ have \ Pythagoras' \ theorem\\\\\\\large (Hypotenuse)^2=(Base)^2+(Perpendicular)^2\\\\\\\large Putting \ value \ here \ we \ get\\\\\\\large (Hypotenuse)^2=(4)^2+(3)^2\\\\\\\large (Hypotenuse)^2=16+9\\\\\\\large (Hypotenuse)^2=25\\\\\\\large (Hypotenuse)=\sqrt{25}\\\\\\\large (Hypotenuse)=5$

$\large For \ cos \ \theta =\dfrac{base}{Hypotenuse} \\\\\\\large Put \ the \ value \ here\\\\\\\large cos \ \theta =\dfrac{4}{5}$

Now put cos θ in given

$\Large \dfrac{1-cos \ \theta}{1+cos \ \theta} \\\\\\\\\Large \text{ \implies \dfrac{1-\dfrac{4}{5}}{1+\dfrac{4}{5}} }\\\\\\\\\Large \text{ \implies \dfrac{\dfrac{5-4}{5}}{\dfrac{5+4}{5}} }\\\\\\\\\Large \text{ \implies \dfrac{\dfrac{1}{5}}{\dfrac{9}{5}} }\\\\\\\\\Large \implies \dfrac{5\times1}{5\times9}\\\\\\\\\Large \implies \dfrac{1}{9}$

Thus we get answer.