Show that (1+ tanθ) divided (1+ cotθ)= tan θ
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.
what a lot of work!
(1 + tanθ)/(1 + cotθ)
= (1 + tanθ)/(1 + 1/tanθ)
= (1 + tanθ)/((tanθ+1)/tanθ)
= tanθ
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.
To prove the given identity (1 + tanθ)/(1 + cotθ) = tanθ, we will start by manipulating the left-hand side (LHS) of the equation:
Step 1: Write the given expression.
(1 + tanθ)/(1 + cotθ)
Step 2: Convert tanθ to sinθ/cosθ and cotθ to cosθ/sinθ.
[(1 + sinθ/cosθ)/(1 + cosθ/sinθ)]
Step 3: Simplify the numerator by finding a common denominator.
[(sinθcosθ + sinθ)/(cosθ)]
Step 4: Multiply the numerator by cosθ/cosθ to simplify.
[(sinθcosθ + sinθ)cosθ/(cosθcosθ)]
Step 5: Distribute cosθ into the numerator.
[sinθcos²θ + sinθcosθ/(cos²θ)]
Step 6: Simplify the numerator by applying the identity cos²θ = 1 - sin²θ.
[(sinθ(1 - sin²θ) + sinθcosθ)/(cos²θ)]
Step 7: Expand and simplify the numerator further.
[sinθ - sin³θ + sinθcosθ/(cos²θ)]
Step 8: Use the identity sinθcosθ = sin2θ/2.
[sinθ - sin³θ + sin2θ/2/(cos²θ)]
Step 9: Combine the terms in the numerator.
[sinθ(1 - sin²θ) + sin2θ/2/(cos²θ)]
Step 10: Simplify (1 - sin²θ) to cos²θ.
[sinθcos²θ + sin2θ/2/(cos²θ)]
Step 11: Cancel out cos²θ in the numerator with the denominator.
[sinθ + sin2θ/2/cos²θ]
Step 12: Write sin2θ in terms of 2sinθcosθ.
[sinθ + (2sinθcosθ)/2cos²θ]
Step 13: Divide the numerator by 2cos²θ.
[sinθ/cos²θ + sinθcosθ/cos²θ]
Step 14: Simplify sinθ/cos²θ to tanθ and sinθcosθ/cos²θ to tanθ.
[tanθ + tanθ]
Step 15: Combine the terms.
2tanθ
So, we have shown that (1 + tanθ)/(1 + cotθ) = tanθ is equivalent to 2tanθ = tanθ. As 2tanθ is equal to tanθ, we have successfully proved the given identity.