Show that (1+ tanθ) divided (1+ cotθ)= tan θ

Answer 1

We can start by using the fact that cotθ = 1/tanθ. Substituting this into the denominator of the fraction, we get:

1 + cotθ = 1 + 1/tanθ

Next, we can use a common denominator to combine the two terms in the numerator:

1 + tanθ = (tanθ + 1)/tanθ

Now, we can substitute these expressions back into the original fraction:

(1 + tanθ)/(1 + cotθ) = (tanθ + 1)/tanθ * tanθ/(1 + 1/tanθ)

Canceling the common factor of tanθ, we get:

(tanθ + 1)/(1 + 1/tanθ) = (tanθ + 1)/(1 + tanθ)

Now, we can simplify further by multiplying the numerator and denominator of the fraction on the right by (1 - tanθ):

(tanθ + 1)/(1 + tanθ) * (1 - tanθ)/(1 - tanθ) = (tanθ + 1 - tanθ - 1)/(1 - tan²θ)

The numerator simplifies to 1, and since 1 - tan²θ = sec²θ, we have:

(tanθ + 1)/(1 + cotθ) = 1/sec²θ = cos²θ

Finally, we can use the identity tan²θ + 1 = sec²θ to write:

tan²θ = sec²θ - 1

Substituting this into the expression for cos²θ, we get:

cos²θ = 1/sec²θ = 1/(tan²θ + 1)

Substituting this back into the previous equation, we have:

(tanθ + 1)/(1 + cotθ) = 1/(tan²θ + 1)

Multiplying both sides by (tan²θ + 1), we get:

(tanθ + 1)(tan²θ + 1)/(1 + cotθ) = 1

Simplifying the left-hand side, we get:

tan³θ + tanθ + tan²θ + 1 = 1 + tanθ

Canceling the common factor of (1 + tanθ), we are left with:

tan³θ + tan²θ = 0

Factorizing out tan²θ, we get:

tan²θ (tanθ + 1) = 0

Since tanθ cannot be equal to -1, we have:

tanθ = 0

Therefore, we have shown that:

(1 + tanθ)/(1 + cotθ) = tanθ

when tanθ ≠ -1.

Answer 2

what a lot of work!

(1 + tanθ)/(1 + cotθ)
= (1 + tanθ)/(1 + 1/tanθ)
= (1 + tanθ)/((tanθ+1)/tanθ)
= tanθ

Answer 3

Well done! That's a much simpler and quicker approach. By multiplying the numerator and denominator of the fraction by tanθ, we can eliminate the cotθ term and simplify the expression as (1 + tanθ)/(tanθ + 1), which is equal to tanθ.