Question

5cos theta+ 12 sin theta = 13 find tan theta?

Answer 1

Topic :-

Trigonometry

Given :-

$\sf{5\cos\theta+12\sin\theta=13}$

To Find :-

$\sf{\tan\theta=?}$

Solution :-

$\sf{\implies5\cos\theta+12\sin\theta=13}$

Divide whole equation with cosθ,

$\sf{\implies 5\cdot\dfrac{\cos\theta}{\cos\theta}+12\cdot\dfrac{\sin\theta}{\cos\theta}=\dfrac{13}{\cos\theta}}$

$\sf{\implies 5+12\tan\theta=13\sec\theta}$

Squaring both sides,

$\sf{\implies (5+12\tan\theta)^2=(13\sec\theta)^2}$

$\sf{\implies 25+2(5)(12\tan\theta)+144\tan^2\theta=169\sec^2\theta}$

$\sf{(\because (a+b)^2=a^2+2ab+b^2)}$

$\sf{\implies 25+120\tan\theta+144\tan^2\theta=169\sec^2\theta}$

$\sf{\implies 25+120\tan\theta+144\tan^2\theta=169(1+\tan^2\theta)}$

$(\because \sec^2\theta=1+\tan^2\theta)$

$\sf{\implies 25+120\tan\theta+144\tan^2\theta=169+169\tan^2\theta}$

$\sf{\implies 0=169\tan^2\theta-144\tan^2\theta-120\tan\theta+169-25}$

$\sf{\implies 25\tan^2\theta-120\tan\theta+144=0}$

Splitting middle term and factorising it,

$\sf{\implies 25\tan^2\theta-60\tan\theta-60\tan\theta+144=0}$

$\sf{\implies 5\tan\theta(5\tan\theta-12)-12(5\tan\theta-12)=0}$

$\sf{\implies (5\tan\theta-12)(5\tan\theta-12)=0}$

$\sf{\implies (5\tan\theta-12)^2=0}$

Taking square root both sides,

$\sf{\implies 5\tan\theta-12=0}$

$\sf{\implies 5\tan\theta=12}$

$\sf{\implies \tan\theta=\dfrac{12}{5}}$

Answer :-

$\underline{\boxed{\sf{\tan\theta=\dfrac{12}{5}}}}$

Additional Formulae :-

$\sf{sin^2\theta+\cos^2\theta=1}$

$\sf{1+\cot^2\theta=\csc^2\theta}$

$\sf{\sin 2\theta=2\sin\theta\cos\theta}$

$\sf{\cos2\theta=\cos^2\theta-\sin^2\theta}$

$\sf{\sin3\theta=3\sin\theta-4\sin^3\theta}$

$\sf{\cos3\theta=4\cos^3\theta-3\cos\theta}$