Question

if tan theta = 4/3 then find the value of 5sin theta + 2 cos theta / 3 sin theta - cos theta ​

Answer 1

Given :-

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

To Find :-

$\sf \dfrac{5 sin \theta + 2 cos \theta}{3 sin \theta - cos \theta}$

Solution :-

We know :-

$\bf\dag \ sec^2 \theta = 1+ tan^2 \theta$

$\sf : \implies sec^2 \theta = 1+ \left( \dfrac{4}{3} \right)^2$

$\sf : \implies sec^2 \theta = 1 + \dfrac{16}{9}$

$\sf : \implies sec^2 \theta = \dfrac{25}{9}$

$\sf : \implies sec \theta = \dfrac{5}{3}$

We know :-

$\bf\dag \ cos \theta = \dfrac{1}{sec\theta}$

$\sf : \implies cos \theta = \dfrac{3}{5}$

We know :-

$\bf\dag \ sin \theta = cos \theta tan \theta$

$\sf : \implies sin \theta = \dfrac{3}{5} \times \dfrac{4}{3}$

$\sf : \implies sin \theta = \dfrac{4}{5}$

$\sf : \implies \dfrac{5 sin \theta + 2 cos \theta }{3 sin \theta - cos \theta }= \dfrac{5 \times \dfrac{4}{5}+ 2 \times \dfrac{3}{5}}{3 \times \dfrac{4}{5}- \dfrac{3}{5}}$

$\sf : \implies \dfrac{5 sin \theta + 2 cos \theta }{ 3sin \theta - cos \theta }= \dfrac{4+\dfrac{6}{5}}{\dfrac{12}{5} - \dfrac{3}{5}}$

$\sf : \implies \dfrac{5sin \theta +2 cos \theta}{3 sin \theta - cos \theta }= \dfrac{\dfrac{(4\times 5 )+6}{5}}{\dfrac{12-3}{5}}$

$\sf : \implies \dfrac{5sin \theta +2 cos \theta}{3 sin \theta - cos \theta }= \dfrac{26}{5} \times \dfrac{5}{9}$

$\bf : \implies \dfrac{5sin \theta +2 cos \theta}{3 sin \theta - cos \theta } = \dfrac{26}{9}$

Answer 2

Glad to help you!

thanks