# Let $0\,^{\circ}\mathrm{} < A < 45\,^{\circ}\mathrm{}$ . If $420(\tan A + \cot A) = 841$ then ﬁnd the value of $(116 \cos A − 58 \sin A)$.

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 )

• Recognize $58(2\cos A-\sin A)$. Try to manipulate this further to get to the tangent. Try to isolate terms that look like the original equation. Multiply the first equation by tangent: $$\frac{420}{\cos^2A}=841\tan A$$ and $$420=841\cos A \sin A$$ I'd continue from here. Mar 30 '14 at 12:45
• @orion I did recognize that right after seeing the problem . I will try to do what you said and post if I get any difficulties , Thanks ! Mar 30 '14 at 12:48
• @orion I didn't get what you did ( After "Multiply the first equation by tangent:") Mar 30 '14 at 13:00
• I used $1+\tan^2 x=\frac{1}{\cos^2 x}$. Mar 30 '14 at 13:08

## 2 Answers

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 $\displaystyle\tan A=\frac{\sin A}{\cos A}\iff\sin A=\tan A\cos A$

• Thank you , but I think this is a tedious way to solve the question because solving that quadratic equation is tedious ( I'm not allowed to use computer or calculator ) and the answer will not be a simple angle ( I'm not allowed to use trignometric tables ) . Instead do you have an idea on how to convert $2\cos A-\sin A$ in terms of $\tan A$? Mar 30 '14 at 12:54
• $\tan A+\frac1{\tan A}=\frac{841}{420} \iff420\tan^2A-841\tan A+240=0$ Mar 30 '14 at 13:05
• @AGoogler, please find the edited version Mar 30 '14 at 13:11
• @zaarcis, you made a typo while rectifying me :) Mar 30 '14 at 13:11
• @labbhattacharjee Ouch, yes. :) Mar 30 '14 at 13:14

All MPA trig questions are meant to be solved using quadratics. That's why the give angle is smaller that 45.

• I think the answer belongs here: math.stackexchange.com/questions/732598/… Besides, which competition are you both giving? Apr 1 '14 at 14:14
• Yep I scanned answer of that and while reading this posted this here. The competition is M. prakash academy qualification test. One of the hardest competition. Academy for IIT traning.
– Ajay
Apr 1 '14 at 14:20