Let $0\,^{\circ}\mathrm{} < A < 45\,^{\circ}\mathrm{}$. If $$420(\tan A + \cot A) = 841$$ then find the value of $$(116 \cos A −
58 \sin A)$$
One way to solve this is by usual method , that is putting $\cot A = 1/\tan A$ in first equation then finding angle A and then calculating the answer.
But here I cannot do that because the quadratic equation will be very complicated ( 420 times two is 840 , just 1 less than 841) and secondly the angle A won't be simple angle and I'm not allowed to use any tables.
So how should I solve this? Any hints are appreciated.
(This is not class-homework , I'm solving sample questions for a competitive exam )
Best Answer
Observe that as $\displaystyle0< A<45^\circ, 0<\tan A<\tan45^\circ=1$
So,we have $\displaystyle\tan A+\frac1{\tan A}=\frac{841}{420}$
$\displaystyle\iff420\tan^2A-841\tan A+420=0$
So, the discriminant will be $\displaystyle841^2-4\cdot420\cdot420=841^2-840^2=841+840=41^2$
Solve the Quadratic Equation for $\tan A$ to get $\displaystyle\tan A=\frac{841\pm41}{420}$
Clearly, $\displaystyle\tan A=\frac{841+41}{2\cdot420}>1$
So, we need $\displaystyle\tan A=\frac{841-41}{2\cdot420}=\cdots$ as it lies in our required range
Also, $\displaystyle\sin A>0, \cos A>0$
So, $\displaystyle\cos A=+\frac1{\sqrt{\sec^2A}}=+\frac1{\sqrt{1+\tan^2A}}$ and $\displaystyle\tan A=\frac{\sin A}{\cos A}\iff\sin A=\tan A\cos A $