If 3 cos θ − 4 sin θ = 2 cos θ + sin θ, find tan θ.

Solution:

Given: $3 \cos \theta-4 \sin \theta=2 \cos \theta+\sin \theta$

To find: $\tan \theta$

Now consider the given expression

$3 \cos \theta-4 \sin \theta=2 \cos \theta+\sin \theta$

Now by dividing both sides of the above expression by $\cos \theta$

We get,

$\frac{3 \cos \theta-4 \sin \theta}{\cos \theta}=\frac{2 \cos \theta+\sin \theta}{\cos \theta}$

Now by separating the denominator for each terms

We get,

$\frac{3 \cos \theta}{\cos \theta}-\frac{4 \sin \theta}{\cos \theta}=\frac{2 \cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}$

Now in the above expression $\cos \theta$ present in both numerator and denominator gets cancelled

Therefore,

$3-\frac{4 \sin \theta}{\cos \theta}=2+\frac{\sin \theta}{\cos \theta}$....(1)

Now we know that,

$\frac{\sin \theta}{\cos \theta}=\tan \theta$

Therefore by substituting $\frac{\sin \theta}{\cos \theta}=\tan \theta$ in equation(1)

We get,

$3-4 \tan \theta=2+\tan \theta$

Now by taking $\tan \theta$ on L.H.S

We get,

$-\tan \theta-4 \tan \theta=2-3$

Therefore,

$-5 \tan \theta=-1$

$5 \tan \theta=1$

$\tan \theta=\frac{1}{5}$

Hence $\tan \theta=\frac{l}{5}$