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 $a\tan \theta + b\sec \theta = c$ ,when $\left| b \right| \leqslant \sqrt {\left( {{a^2} + {c^2}} \right)} $ ,then ${a^2} - {c^2}$ is equal to
$
  {\text{A}}{\text{. }}ac \\
  {\text{B}}{\text{. 2}}ac \\
  {\text{C}}{\text{. }}\dfrac{a}{c} \\
  {\text{D}}{\text{. }}None{\text{ }}of{\text{ }}these \\
 $

Hint: To solve this question we have to assume the other two angles and solve the other equation given taking the two angles as their roots and then by combining the solutions of both the equations we find the correct answer.

Complete step-by-step answer:
Considering the question let the two values of $\theta $ be ${\theta _1}$ and ${\theta _2}$ .Therefore, by the property of sum of internal angles of a triangle ${\theta _1} + {\theta _2} = \dfrac{\pi }{4}$ ……$\left( 1 \right)$

Now, from the equation given in question we get

$

  b\sec \theta = c - a\tan \theta \\

   \Rightarrow \sec \theta = \dfrac{c}{b} - \dfrac{a}{b}\tan \theta \\

 $

Squaring on both sides and simplifying it gives

$

  {\sec ^2}\theta = {\left( {\dfrac{c}{b} - \dfrac{a}{b}\tan \theta } \right)^2} \\

   \Rightarrow {\sec ^2}\theta = \dfrac{{{c^2}}}{{{b^2}}} + \dfrac{{{a^2}}}{{{b^2}}}{\tan ^2}\theta - \dfrac{{2ac}}{{{b^2}}}\tan \theta \\

 $

Now, as you know ${\sec ^2}\theta - {\tan ^2}\theta = 1$

$

   \Rightarrow 1 + {\tan ^2}\theta = \dfrac{{{c^2}}}{{{b^2}}} + \dfrac{{{a^2}}}{{{b^2}}}{\tan ^2}\theta - \dfrac{{2ac}}{{{b^2}}}\tan \theta \\

   \Rightarrow {\tan ^2}\theta \left( {\dfrac{{{a^2}}}{{{b^2}}} - 1} \right) -

\dfrac{{2ac}}{{{b^2}}}\tan \theta + \dfrac{{{c^2}}}{{{b^2}}} - 1 = 0 \\

 $

Take $\dfrac{1}{{{b^2}}}$ as common in whole,

$

   \Rightarrow \dfrac{1}{{{b^2}}}\left[ {\left( {{a^2} - {b^2}} \right){{\tan }^2}\theta - 2ac\tan \theta + {c^2} - {b^2}} \right] = 0 \\

   \Rightarrow \left( {{a^2} - {b^2}} \right){\tan ^2}\theta - 2ac\tan \theta + {c^2} - {b^2} = 0 \\

 $

As roots of this equation is $\tan {\theta _1}$ and $\tan {\theta _2}$

So, by using the concept of sum of roots and product of roots we solve further i.e.

Sum of roots $\left( {\tan {\theta _1} + \tan {\theta _2}} \right) = \dfrac{{ - b}}{a}$ where,

$a{\text{ and }}b$ are coefficients of ${\tan ^2}\theta {\text{ and tan}}\theta $ respectively.

$

  \therefore \tan {\theta _1} + \tan {\theta _2} = \dfrac{{2ac}}{{{a^2} - {b^2}}} \\

    \\

 $

And product of roots $\left( {\tan {\theta _1}.\tan {\theta _2}} \right) = \dfrac{c}{a}$ where, $a$ is a coefficients of ${\tan ^2}\theta {\text{ }}$ are $c$ is a constant.

$\tan {\theta _1}.\tan {\theta _2} = \dfrac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}}$

Taking $\tan $ on both sides of the equation no $\left( 1 \right)$

$\tan \left( {{\theta _1} + {\theta _2}} \right) = \tan \dfrac{\pi }{4}$

Using the formula for $\tan \left( {{\theta _1} + {\theta _2}} \right)$ and since $\tan \dfrac{\pi }{4} = 1$ we get

$\dfrac{{\tan {\theta _1} + \tan {\theta _2}}}{{1 - \tan {\theta _1}.\tan {\theta _2}}} = 1$
Substitute, the values of $\tan {\theta _1}{\text{ and }}\tan {\theta _2}$ from the equations of sum and product of roots

$

   \Rightarrow \dfrac{{\dfrac{{2ac}}{{{a^2} - {b^2}}}}}{{1 - \dfrac{{{c^2} - {b^2}}}{{{a^2} - {b^2}}}}} = 1 \\

   \Rightarrow \dfrac{{2ac}}{{{a^2} - {b^2} - {c^2} + {b^2}}} = 1 \\

   \Rightarrow \dfrac{{2ac}}{{{a^2} - {c^2}}} = 1 \\

   \Rightarrow {a^2} - {c^2} = 2ac \\

 $

Therefore answer is ${\text{B}}{\text{. 2}}ac$

Note: While solving this question one have to find the sum and product of roots of the given quadratic equation and we don’t need the values of the roots and students need to remember the trigonometric formulas required at each step and one formula of geometry is also used which is the sum of internal angles of a triangle is ${180^ \circ }{\text{ i}}{\text{.e}}{\text{.}}\pi $. Now, solving the equations properly step by step is also important and substituting the sum and product values will guarantee the correct answer.