cos ( x - y ) / cos x cos y = 1 + tan x tan y is proven as explained below.
Further explanation
Firstly , let us learn about trigonometry in mathematics.
Suppose the ΔABC is a right triangle and ∠A is 90°.
sin ∠A = opposite / hypotenuse
cos ∠A = adjacent / hypotenuse
tan ∠A = opposite / adjacent
There are several trigonometric identities that need to be recalled, i.e.
[tex]cosec ~ A = \frac{1}{sin ~ A}[/tex]
[tex]sec ~ A = \frac{1}{cos ~ A}[/tex]
[tex]cot ~ A = \frac{1}{tan ~ A}[/tex]
[tex]tan ~ A = \frac{sin ~ A}{cos ~ A}[/tex]
Let us now tackle the problem!
In this problem , we will use identity as follows:
[tex]\large {\boxed {\cos (x - y) = \cos x ~ \cos y + \sin x ~ \sin y } }[/tex]
Given:
[tex](\cos ( x - y )) / (\cos x \cos y) = \frac{(\cos x ~ \cos y + \sin x ~ \sin y)}{(\cos x \cos y)}[/tex]
[tex](\cos ( x - y )) / (\cos x \cos y) = \frac{(\cos x ~ \cos y)}{(\cos x \cos y)} + \frac{(\sin x ~ \sin y)}{(\cos x \cos y)}[/tex]
[tex](\cos ( x - y )) / (\cos x \cos y) = 1 + \frac{(\sin x ~ \sin y)}{(\cos x \cos y)}[/tex]
[tex](\cos ( x - y )) / (\cos x \cos y) = 1 + \frac{(\sin x)}{(\cos x)} \frac{(\sin y)}{(\cos y)}[/tex]
[tex]\large {\boxed {(\cos ( x - y )) / (\cos x \cos y) = 1 + \tan x \tan y} }[/tex]
Learn more
- Calculate Angle in Triangle : https://brainly.com/question/12438587
- Periodic Functions and Trigonometry : https://brainly.com/question/9718382
- Trigonometry Formula : https://brainly.com/question/12668178
Answer details
Grade: College
Subject: Mathematics
Chapter: Trigonometry
Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse , Triangle , Fraction , Lowest , Function , Angle
