Question

The number of solutions of the equation ${{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\dfrac{\pi }{4}$ is
a. 2
b. 3
c. 1
d. None of these

Answer 1

Hint:In order to find the solution of this question, we should have some knowledge about the inverse trigonometric formulas like ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. Also, we should know that $\tan \dfrac{\pi }{4}=1$. By using these formulas we can solve the question.

Complete step-by-step answer:
In this question, we have been asked to find the number of solutions of the equation ${{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\dfrac{\pi }{4}$. To solve this equation, we should know that ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. So, for a = 2x and b = 3x, we can write ${{\tan }^{-1}}\left( \dfrac{2x+3x}{1-\left( 2x \right)\left( 3x \right)} \right)=\dfrac{\pi }{4}$. Now, we can further simplify it as,
${{\tan }^{-1}}\left( \dfrac{5x}{1-6{{x}^{2}}} \right)=\dfrac{\pi }{4}$
Now, we know that tangent ratios of an equality are equal. So, we will take tangent ratio of the equality. Therefore, we will get,
$\tan \left( {{\tan }^{-1}}\left( \dfrac{5x}{1-6{{x}^{2}}} \right) \right)=\tan \left( \dfrac{\pi }{4} \right)$
Now, we know that $\tan \left( {{\tan }^{-1}}x \right)=x$, so we can write the equation as,
$\dfrac{5x}{1-6{{x}^{2}}}=\tan \left( \dfrac{\pi }{4} \right)$
We know that $\tan \left( \dfrac{\pi }{4} \right)=1$, so we can write the equation as,
$\dfrac{5x}{1-6{{x}^{2}}}=1$
Now, we will cross multiply the equation. So, we will get,
$5x=1-6{{x}^{2}}$
We can also write it as,
$6{{x}^{2}}+5x-1=0$
Now, we know that the roots of a quadratic equation, $a{{x}^{2}}+bx+c=0$ can be calculated using the formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. So, for the equation $6{{x}^{2}}+5x-1=0$, we have a = 6, b = 5 and c = -1. So, we get the value of x as,
$x=\dfrac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 6 \right)\left( -1 \right)}}{2\left( 6 \right)}$
We can further write it as,

& x=\dfrac{-5\pm \sqrt{25+24}}{12} \\\ & \Rightarrow \dfrac{-5\pm \sqrt{49}}{12} \\\ & \Rightarrow \dfrac{-5\pm 7}{12} \\\ & \Rightarrow \dfrac{-5+7}{12},\dfrac{-5-7}{12} \\\ & \Rightarrow \dfrac{2}{12},\dfrac{-12}{12} \\\ & \Rightarrow \dfrac{1}{6},-1 \\\ & \therefore x=\dfrac{1}{6},-1 \\\ \end{aligned}$$ Hence, we get that there are 2 possible solutions for the equation, ${{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\dfrac{\pi }{4}$. Therefore, option (a) is the correct answer. Note: We can also solve this question using the middle term split method while solving the quadratic equation and then find the value of x from it. The students can use any method to solve this question, but it would be always better to choose the method which is less time consuming and the method which the students know properly. Also we have to remember that ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$.