Question

If in an obtuse angled triangle the obtuse angle is \[\dfrac{3\pi }{4}\] and the other two angles are equal to two values of \[\theta \] satisfying \[a\tan \theta +b\sec \theta =c\], where \[|b|\le \sqrt{{{a}^{2}}+{{c}^{2}}}\], then find the value of \[{{a}^{2}}-{{c}^{2}}\]

Answer 1

Hint: We will transform \[b\sec \theta =c-a\tan \theta \] in quadratic equation and then we will use sum of roots formula and product of roots formula to solve this question. Also one angle is given as \[\dfrac{3\pi }{4}\], so we can find the sum of the other two angles as the sum of the angles of a triangle is 180.

Complete step-by-step answer:

It is mentioned in the question that in an obtuse angled triangle, the obtuse angle is \[\dfrac{3\pi }{4}\] and other two angles are equal to two values of \[\theta \] satisfying \[a\tan \theta +b\sec \theta =c\].

\[a\tan \theta +b\sec \theta =c.......(1)\]

Rearranging equation (1) we get,

\[b\sec \theta =c-a\tan \theta .......(2)\]

Squaring both sides of equation (2) we get,

\[{{b}^{2}}{{\sec }^{2}}\theta ={{c}^{2}}+{{a}^{2}}{{\tan }^{2}}\theta -2ac\tan \theta .......(3)\]

Substituting \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \] in equation (3) we get,

\[{{b}^{2}}+{{b}^{2}}{{\tan }^{2}}\theta ={{c}^{2}}+{{a}^{2}}{{\tan }^{2}}\theta -2ac\tan \theta .......(4)\]

Simplifying and rearranging equation (4) we get,

\[({{a}^{2}}-{{b}^{2}}){{\tan }^{2}}\theta -2ac\tan \theta -({{b}^{2}}-{{c}^{2}})=0.......(5)\]

Now comparing equation (5) with the quadratic equation \[a{{x}^{2}}+bx+c=0\] we get\[x=\tan \theta \], a as \[{{a}^{2}}-{{b}^{2}}\], b as \[-2ac\] and c as \[{{c}^{2}}-{{b}^{2}}\].

Let the two roots of equation (5) be \[\tan {{\theta }_{1}}\] and \[\tan {{\theta }_{2}}\].

Sum of roots \[=\tan {{\theta }_{1}}+\tan {{\theta }_{2}}=\dfrac{-b}{a}=\dfrac{2ac}{{{a}^{2}}-{{b}^{2}}}.......(6)\]

Product of roots \[=\tan {{\theta }_{1}}\tan {{\theta }_{2}}=\dfrac{c}{a}=\dfrac{{{c}^{2}}-{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}.......(7)\]

Now we know that sum of the three angles in a triangle is equal to 180. So using this information we get,\[{{\theta }_{1}}+{{\theta }_{2}}+\dfrac{3\pi }{4}=\pi .......(8)\]

Solving equation (8) we get,

\[\begin{align}

  & {{\theta }_{1}}+{{\theta }_{2}}=\pi -\dfrac{3\pi }{4} \\

 & {{\theta }_{1}}+{{\theta }_{2}}=\dfrac{\pi }{4}..........(9) \\

\end{align}\]

Now applying tan on both sides of equation (9) we get,

\[\tan ({{\theta }_{1}}+{{\theta }_{2}})=\tan \dfrac{\pi }{4}..........(10)\]

We know that \[\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\], so applying this formula in equation (10) we get,

\[\dfrac{\tan {{\theta }_{1}}+\tan {{\theta }_{2}}}{1-\tan {{\theta }_{1}}\tan {{\theta }_{2}}}=1..........(11)\]

Now substituting sum of roots from equation (6) and product of roots from equation (7) in equation (11) we get,

\[\dfrac{\dfrac{2ac}{{{a}^{2}}-{{b}^{2}}}}{1-\dfrac{{{c}^{2}}-{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}}=1..........(12)\]

Now rearranging and simplifying equation (12) we get,

\[\dfrac{\dfrac{2ac}{{{a}^{2}}-{{b}^{2}}}}{\dfrac{{{a}^{2}}-{{b}^{2}}-{{c}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}}=1..........(13)\]

Cancelling similar terms in equation (13) we get,

\[\dfrac{2ac}{{{a}^{2}}-{{c}^{2}}}=1..........(14)\]

Solving for \[{{a}^{2}}-{{c}^{2}}\] in equation (14) we get,

\[\Rightarrow {{a}^{2}}-{{c}^{2}}=2ac\]

Hence the value of \[{{a}^{2}}-{{c}^{2}}\] is \[2ac\].


Note: Remembering the concept of quadratic equation and the formulas of sum of roots and product of roots is important. Also we need to remember that \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \] and \[\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\]. In a hurry we can make a mistake in solving equation 12 so we need to be careful while doing this step.