Obtuse-angled triangle equation

Question

Give a obtuse-angled triangle and the obtuse angle is 105º. Find n such that the acute angles be the roots of the equation.

$$3\sec(x)+n\left(\frac{1}{\sec(x)}-\frac{1}{\csc(x)}\right)=3\left(\frac{1}{\sec(x)}+\frac{1}{\csc(x)}\right)$$

Answer 1

If $A,B$ are the rest two acute angles, we immediately have $A+B=(180-105)^\circ=75^\circ$

On simplification of the given expression, $$\frac3{\cos x}=3(\cos x+\sin x)-n(\cos x-\sin x)$$ $$\implies \frac3{\cos x}=\sin x(3+n)+\cos x(3-n)\ \ \ \ (1)$$

$$\implies 3=(3+n)\sin x\cos x+(3-n)\cos^2x \ \ \ \ (2)$$

Multiplying either sides by $\sec^2x=1+\tan^2x,$

$$\implies 3(1+\tan^2x)=(3+n)\tan x+(3-n) \ \ \ \ (3)$$

We could directly go to $(3)$ from $(1)$ dividing either sides by $\cos x,$ but the terms in $(2)$ are closely related to Double-Angle Formulas which immediately reminds me of Tangent half-angle formulae

Now, rearrange $(3)$ to get a Quadratic Equation in $\tan x$ whose two roots are $\tan A,\tan B$

Now use Vieta's formulas to find $\displaystyle\tan A+\tan B$ and $\displaystyle\tan A\cdot\tan B$

and then use Angle addition formula of tangents in $\displaystyle\tan75^\circ=\tan(A+B)=\cdots$