Tan theta / 1- tan theta -cot theta /1-cot theta =cos theta + sin theta/cos theta -sin theta
Brackets are essential here, you probably meant:
TanØ / (1- tan Ø) - cot Ø/(1-cot Ø) = cos Ø + sin Ø/(cos Ø - sin Ø)
check this equation.
What are we doing here, solving? or proving it as an identity?
BTW, I tried it with Ø = 20° , and LS ≠ RS
TanØ / (1- tan Ø) - cot Ø/(1-cot Ø) = (cos Ø + sin Ø)/(cos Ø - sin Ø)
TanØ / (1- tan Ø)
multiply top and bottom by cosØ to get sinØ/(cosØ-sinØ)
similarly, cotØ/(1-cotØ) = cosØ/(sinØ-cosØ)
Now just add them together ...
To prove the given trigonometric identity, we need to simplify the left-hand side (LHS) and right-hand side (RHS) separately and then show that they are equal.
Let's start by simplifying the LHS:
LHS = (tan(theta) / (1 - tan(theta))) - (cot(theta) / (1 - cot(theta)))
First, we can replace tan(theta) with sin(theta) / cos(theta) and cot(theta) with cos(theta) / sin(theta):
LHS = (sin(theta) / cos(theta)) / (1 - sin(theta) / cos(theta)) - (cos(theta) / sin(theta)) / (1 - cos(theta) / sin(theta))
Now, let's find the least common denominator (LCD) for the fractions in both terms:
The LCD is cos(theta) for the first term and sin(theta) for the second term.
Multiplying the numerator and denominator of the first term by cos(theta) and the numerator and denominator of the second term by sin(theta), we get:
LHS = (sin(theta) / cos(theta)) * (cos(theta) / cos(theta)) / ((cos(theta) - sin(theta)) / cos(theta)) - (cos(theta) / sin(theta)) * (sin(theta) / sin(theta)) / ((sin(theta) - cos(theta)) / sin(theta))
Simplifying further:
LHS = sin(theta) / (cos(theta) - sin(theta)) - cos(theta) / (sin(theta) - cos(theta))
Now, let's simplify the RHS:
RHS = (cos(theta) + sin(theta)) / (cos(theta) - sin(theta))
Now, to establish the identity, we need to show that LHS = RHS.
To do that, let's simplify the LHS further:
LHS = sin(theta) / (cos(theta) - sin(theta)) - cos(theta) / (sin(theta) - cos(theta))
Combining the fractions by finding a common denominator, we get:
LHS = (sin(theta) * (sin(theta) - cos(theta)) - cos(theta) * (cos(theta) - sin(theta))) / ((cos(theta) - sin(theta)) * (sin(theta) - cos(theta)))
Expanding and simplifying, we get:
LHS = (sin^2(theta) - sin(theta) * cos(theta) - cos^2(theta) + cos(theta) * sin(theta)) / (cos^2(theta) - 2 * cos(theta) * sin(theta) + sin^2(theta))
Canceling out the common terms, we have:
LHS = (sin^2(theta) - cos^2(theta)) / (cos^2(theta) - sin^2(theta))
Using the trigonometric identity sin^2(theta) + cos^2(theta) = 1, we can simplify further:
LHS = (1 - cos^2(theta) - cos^2(theta)) / (cos^2(theta) - (1 - cos^2(theta)))
Simplifying:
LHS = (1 - 2 cos^2(theta)) / (2 cos^2(theta) - 1)
Now, let's simplify the RHS:
RHS = (cos(theta) + sin(theta)) / (cos(theta) - sin(theta))
We can multiply the numerator and denominator by (cos(theta) + sin(theta)) to simplify further:
RHS = (cos(theta) + sin(theta)) * (cos(theta) + sin(theta)) / ((cos(theta) - sin(theta)) * (cos(theta) + sin(theta)))
Expanding and simplifying, we get:
RHS = (cos^2(theta) + 2 cos(theta) * sin(theta) + sin^2(theta)) / (cos^2(theta) - sin^2(theta))
Using the trigonometric identity sin^2(theta) + cos^2(theta) = 1, we can further simplify:
RHS = (1 + 2 cos(theta) * sin(theta)) / (cos^2(theta) - (1 - cos^2(theta)))
Simplifying:
RHS = (1 + 2 cos(theta) * sin(theta)) / (2 cos^2(theta) - 1)
Now, we can see that the LHS = RHS:
LHS = (1 - 2 cos^2(theta)) / (2 cos^2(theta) - 1)
RHS = (1 + 2 cos(theta) * sin(theta)) / (2 cos^2(theta) - 1)
Therefore, the given trigonometric identity is valid.