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Last updated date: 30th Nov 2023
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# Prove that tan $\left( {x - \left. y \right)} \right.$$= \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$

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.
We know, tan $\theta$ $\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$
tan $\left( {x - \left. y \right)} \right.$ $= \dfrac{{\sin \left( x \right. - \left. y \right)}}{{\cos \left( {x - \left. y \right)} \right.}}$
$\sin \left( {x - y} \right) = \sin x\cos y - \sin y\cos x \\ \cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y \\ \tan \left( {x - y} \right) = \dfrac{{\sin x\cos y - \sin y\cos x}}{{\cos x\cos y + \sin x\sin y}}$
$\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$ 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.