Question

If $7\sin A+24\cos A=25$, find the value of $\tan A$

Answer 1

$7\sin A+24\cos A=25$
squaring both sides we get
$\Rightarrow {(7\sin A+24\cos A)}^2=625\Rightarrow 49{\sin }^{2}A+576{\cos }^{2}A+2\left(7\sin A\right)\left(24\cos A\right)=625[\because {(a+b)}^2=a^2+b^2+2ab]$
divide by ${\cos }^{2}A$ on both sides
$\Rightarrow \frac{49{\sin }^{2}A}{{\cos }^{2}A}+\frac{576{\cos }^{2}A}{{\cos }^{2}A}+\frac{336\sin A.\cos A}{{\cos }^{2}A}=\frac{625}{{\cos }^{2}A}\Rightarrow 49{\tan }^{2}A+576+336\tan A=625{\sec }^{2}A$
$\Rightarrow 49{\tan }^{2}A+576+336\tan A=625(1+{\tan }^{2}A)[\because (1+{\tan }^{2}A)={\sec }^{2}A]$
$\Rightarrow 49{\tan }^{2}A+576+336\tan A=625+625{\tan }^{2}A\Rightarrow 576{\tan }^{2}A-336\tan A+49=0\Rightarrow 576{\tan }^{2}A-168\tan A-168\tan A+49=0$
$\Rightarrow 24\tan A(24\tan A-7)-7(24\tan A-7)=0\Rightarrow {(24\tan A-7)}^2=0\Rightarrow \tan A=\frac{7}{24}$