Question

If $4cos\Theta +3sin\Theta =5,$ find the value of tan$\Theta$

Answer 1

We have,

$4\cos \theta +3\sin \theta =5$

Then,

$4\cos \theta +3\sin \theta =5$

$4\cos \theta +3\sqrt{1-{\cos }^2\theta }=5$

$3\sqrt{1-{\cos }^2\theta }=5-4\cos \theta$

Squaring both side and we get,

$9(1-{\cos }^2\theta )={(5-4\cos \theta )}^2$

$\Rightarrow 9-9{\cos }^2\theta =25+16{\cos }^2\theta -40\cos \theta$

$\Rightarrow 25+16{\cos }^2\theta -40\cos \theta -9+9{\cos }^2\theta =0$

$\Rightarrow 25{\cos }^2\theta -40\cos \theta +16=0$

$\Rightarrow {\left(5\cos \theta \right)}^2-2\times 5\cos \theta \times 4+4^2=0$

$\Rightarrow {(5\cos \theta -4)}^2=0$

$\Rightarrow 5\cos \theta -4=0$

$\Rightarrow 5\cos \theta =4$

$\Rightarrow \cos \theta =\frac{4}{5}$

Put the value of given equation and we get,

$4\cos \theta +3\sin \theta =5$

$\Rightarrow 4\times \frac{4}{5}+3\sin \theta =5$

$\Rightarrow \frac{16}{5}-5=3\sin \theta$

$\Rightarrow 3\sin \theta =\frac{16-25}{5}$

$\Rightarrow \sin \theta =-\frac{9}{15}$

$\Rightarrow \sin \theta =\frac{-3}{5}$

Then,

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

$\tan \theta =\frac{-\frac{3}{5}}{\frac{4}{5}}$

$\tan \theta =\frac{-3}{4}$

Hence, this is the answer.