Question
If in an obtuse-angled triangle the obtuse angle is $$\frac{{3\pi }}{4}$$ and the other two angles are equal to two values of $$\theta $$ satisfying $$a\tan \theta + b\sec \theta = c,$$ where $$\left| b \right| \leqslant \sqrt {{a^2} + {c^2}} ,\,$$ then $${{a^2} - {c^2}}$$ is equal to
A.
$$ac$$
B.
$$2ac$$
C.
$$\frac{a}{c}$$
D.
None of these
Answer :
$$2ac$$
Solution :
$$c\cos \theta - a\sin \theta = b.$$ Therefore, $$c\cos\alpha - a\sin \alpha = c\cos \beta - a\sin \beta ,$$ where $$\alpha ,\beta $$ are the other two angles of the triangle.
$$\eqalign{
& \therefore \,\,c\left( {\cos \alpha - \cos \beta } \right) = a\left( {\sin \alpha - \sin \beta } \right) \cr
& {\text{or, }}\frac{c}{a} = \frac{{2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}}}{{2\sin \frac{{\alpha + \beta }}{2}\sin \frac{{\beta - \alpha }}{2}}} = - \cot \frac{{\alpha + \beta }}{2} \cr
& \therefore \,\,\tan \frac{{\alpha + \beta }}{2} = \frac{{ - a}}{c} \cr
& \therefore \,\,\tan \left( {\alpha + \beta } \right) = \frac{{2 \times \left( {\frac{{ - a}}{c}} \right)}}{{1 - \frac{{{a^2}}}{{{c^2}}}}} = \frac{{2ac}}{{{a^2} - {c^2}}} \cr
& \therefore \,\,\tan \left( {\pi - \frac{{3\pi }}{4}} \right) = \frac{{2ac}}{{{a^2} - {c^2}}}\,\,\,\,{\text{or, }}{a^2} - {c^2} = 2ac. \cr} $$