(tanÈ+cotÈ)(sinÈ+cosÈ)=secÈ+cosecÈ
To prove the given trigonometric identity, we'll start with the left-hand side (LHS) of the equation and simplify it step by step until we reach the right-hand side (RHS) of the equation.
LHS: (tanÈ + cotÈ)(sinÈ + cosÈ)
First, let's recall the definitions of the trigonometric functions involved:
- tanÈ is defined as sinÈ/cosÈ
- cotÈ is defined as cosÈ/sinÈ
- secÈ is defined as 1/cosÈ
- cosecÈ is defined as 1/sinÈ
Now, let's substitute the expressions of tanÈ and cotÈ into the LHS:
LHS = (sinÈ/cosÈ + cosÈ/sinÈ)(sinÈ + cosÈ)
Next, let's simplify the equation by finding a common denominator for the fractions:
LHS = [(sinÈ * sinÈ + cosÈ * cosÈ) / (cosÈ * sinÈ)] * (sinÈ + cosÈ)
Using the Pythagorean identity sinÈ^2 + cosÈ^2 = 1, we can simplify the numerator:
LHS = [1 / (cosÈ * sinÈ)] * (sinÈ + cosÈ)
Simplifying the denominator:
LHS = (sinÈ + cosÈ) / (cosÈ * sinÈ)
Now, let's simplify the RHS of the equation:
RHS = secÈ + cosecÈ
Substituting the definitions of secÈ and cosecÈ:
RHS = (1/cosÈ) + (1/sinÈ)
To add the fractions, we need a common denominator:
RHS = [(1 * sinÈ + 1 * cosÈ) / (cosÈ * sinÈ)]
Combining the terms in the numerator:
RHS = (sinÈ + cosÈ) / (cosÈ * sinÈ)
Now that we have simplified both the LHS and RHS of the equation, we can see that they are equal:
LHS = RHS
Therefore, we have successfully proven the given trigonometric identity:
(tanÈ + cotÈ)(sinÈ + cosÈ) = secÈ + cosecÈ.