Question

If $\tan x\centerdot \tan y\ =\ a$ and $x+y=\dfrac{\pi }{6}$, then $\tan x$ and $\tan y$ satisfy the equation.

Answer 1

Hint: When an equation is given in terms of sine, cosine, tangent. We must use any of the trigonometric identities to make the equation solvable. There are many interrelations between sine, cosine, tan, secant, there are interrelations called identities where you see conditions such that $\theta \in R$, that means inequality is true for all angles so directly think of identity which will make your work easy.

${{\sin }^{2}}\theta +{{\cos }^{2}}\ =\ 1$ for all $\theta $, $\tan \left( A+B \right)\ =\ \dfrac{\tan A+\tan B}{1-\tan A\tan B}$

Complete step-by-step answer:
An equality with sine, cosine or tangent in them is called trigonometric equality. These are solved by the same interrelations known beforehand.

All the interrelations which relate sine, cosine, tangent, secant, cotangent and cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.

Given in the question the relation between x and y:

$x+y=\dfrac{\pi }{6}$

By applying \[\tan \]on both sides of equation, we get:

$\tan \left( x+y \right)=\ \tan \left( \dfrac{\pi }{6} \right)\ =\ \dfrac{1}{\sqrt{3}}$

By general knowledge of trigonometry, we get the formula:

$\tan \left( x+y \right)\ =\ \dfrac{\tan x+\tan y}{1-\tan x\tan y}$

By substituting this equation and doing cross multiplication,

$\sqrt{3}\left( \tan x+\tan y \right)\ =\ 1-\tan x\tan y$

By this equation, by substituting value of $\tan x\centerdot \tan y$, we get:

$\tan x+\tan y\ =\ \dfrac{1-a}{\sqrt{3}}$.

A quadratic equation with product of roots b and sum of roots a is given by \[{{x}^{2}}-ax+b\

=\ 0\]. By this we get equation with, $\tan x,\tan y$ roots are,

\[{{x}^{2}}-\dfrac{\left( 1-a \right)x}{\sqrt{3}}+a\ =\ 0\]

So, the equation will be \[\sqrt{3}{{x}^{2}}-\left( a-1 \right)x+a\sqrt{3}\ =\ 0\]

Therefore, the above quadratic equation has roots $\tan x,\tan y$.

Note: Be careful while substituting the sum of roots don’t forget to take the minus sign in the equation.