Question

Prove that tan \[\left( {x - \left. y \right)} \right.\]\[ = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}\]

Answer 1

Hint: In trigonometry when it comes to a right-angle triangle; there are many formulas in trigonometry but there are few most important basic formulas . The Cos theta or cos θ is the ratio of the adjacent side to the hypotenuse, where θ is one of the acute angles. \[Cos\theta = \dfrac{{Adjacent}}{{Hypotenuse}}\]. While we can find sine value for any angle, there are some angles that are more frequently used in trigonometry.

Complete step-by-step solution:
We know, tan \[\theta \] \[\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}\]
tan \[\left( {x - \left. y \right)} \right.\] \[ = \dfrac{{\sin \left( x \right. - \left. y \right)}}{{\cos \left( {x - \left. y \right)} \right.}}\]
Now using the formulae,
\[
  \sin \left( {x - y} \right) = \sin x\cos y - \sin y\cos x \\
  \cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y \\
  \tan \left( {x - y} \right) = \dfrac{{\sin x\cos y - \sin y\cos x}}{{\cos x\cos y + \sin x\sin y}}\]
Dividing both numerator both denominator by cos x cos y,
\[\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}\] proved.

Note: The tangent is defined as the ratio of the opposite side to the adjacent side. The unit circle definition is $\tan\left(\theta\right)=\dfrac{y}{x}$ or $\tan\left(\theta\right)=\dfrac{\sin\left(\theta\right)}{\cos\left(\theta\right)}$ in right triangle trigonometry (for acute angles only), The tangent function is negative whenever sine or cosine, but not both, are negative: the second and fourth quadrants.