Question

If $acos\theta +bsin\theta =c$ and
$aco{s}^2\theta +2asin\theta cos\theta +bsi{n}^2\theta =c$
where a,b,c, are all unequal quantities, then eliminate $\theta$

Answer 1

$a\cos \theta +b\sin \theta =c$ ...(1)

$a{\cos }^2\theta +b{\sin }^2\theta +2a\sin \theta \cos \theta =c$

$a\frac{(1+\cos 2\theta )}{2}+b\frac{(1-\cos 2\theta )}{2}+a\sin 2\theta =c$

$a+a\cos 2\theta +b-b\cos 2\theta +2a\sin 2\theta =2c$

$(a-b)\cos 2\theta +2a\sin 2\theta =2c-(a+b)$ ...(2)

Squaring (1) on both sides

$a^2{\cos }^2\theta +b^2{\sin }^2\theta +ab\sin 2\theta =c^2$

$a^2\frac{(1+\cos 2\theta )}{2}+b^2\frac{(1-\cos 2\theta )}{2}+ab\sin 2\theta =c^2$

$(a^2-b^2)\cos 2\theta +2ab\sin 2\theta =2c^2-(a^2+b^2)$ ....(3)

From equation (2) $\sin 2\theta =\frac{[2c-(a+b\left)\right]-\left[\right(a-b\left)\cos 2\theta \right]}{2a}$

putting in (3)

$(a^2-b^2)\cos 2\theta +b\left[\right(2c-(a+b))-((a-b)\cos 2\theta \left)\right]=2c^2-a^2-b^2$

$(a^2-b^2)\cos 2\theta +b[2c-a-b-(a-b\left)\cos 2\theta \right]=2c^2-a^2-b^2$

$(a^2-b^2)\cos 2\theta +2bc-ab-b^2-b(a-b)\cos 2\theta =2c^2-a^2-b^2$

$a^2\cos 2\theta -b^2\cos 2\theta +2bc-ab-ab\cos 2\theta +b^2\cos 2\theta =2c^2-a^2$

$(a^2-ab)\cos 2\theta =2c^2-a^2+ab-2bc$

$cos2\theta =\frac{2c^2-a^2+ab-2bc}{a^2-ab}=\frac{2c^2-2bc-(a^2-ab)}{a^2-ab}$

$\cos 2\theta =\frac{2c(c-b)}{a(a-b)}-1$

$\theta =\frac{1}{2}{\cos }^{-1}(\frac{2c(c-b)}{a(a-b)}-1)$