Question

if 3 cot theta = 4, show that ( 4 cos theta - sin theta)/(2cos theta + sin theta) = 4/5​

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

$If\:3cot\theta = 4\:then\:prove$

$\dfrac{4cos\theta-sin\theta}{2cos\theta+sin\theta}=\dfrac{13}{11}$

Given,

$3cot\theta = 4$

To Show :-

$\dfrac{4cos\theta-sin\theta}{2cos\theta+sin\theta}=\dfrac{13}{11}$

Solution :-

$As,\: 3cot\theta = 4$

$cot\theta=\dfrac{4}{3}$

We know that :-

$cot\theta=\dfrac{adjacent\:side}{opposite\:side}$

Adjacent side = 4 , opposite side = 3

By applying pythagoras theorem We will get hypotenuse side = 5

$\implies sin\theta=\dfrac{3}{5}$

$cos\theta=\dfrac{4}{5}$

$= \dfrac{4cos\theta-sin\theta}{2cos\theta+sin\theta}$

$=\dfrac{\left(4\times\dfrac{4}{5}\right)-\dfrac{3}{5}}{\left(2\times\dfrac{4}{5}\right)+\dfrac{3}{5}}$

$=\dfrac{\dfrac{16}{5}-\dfrac{3}{5}}{\dfrac{8}{5}+\dfrac{3}{5}}$

$=\dfrac{\dfrac{16-3}{5}}{\dfrac{8+3}{5}}$

$=\dfrac{\dfrac{13}{5}}{\dfrac{11}{5}}$

$=\dfrac{13}{11}$

taking L.H.S :-