prove that tan(45+theta)+tan(45-theta)=2sec2theta
To prove the equation tan(45+θ) + tan(45-θ) = 2sec^2θ, we will use the following trigonometric identities:
1. tan(A+B) = (tanA + tanB) / (1 - tanA * tanB)
2. tan(A-B) = (tanA - tanB) / (1 + tanA * tanB)
3. sec^2(A) = 1 + tan^2(A)
Using these identities, let's simplify the left-hand side of the equation:
tan(45+θ) + tan(45-θ) = [(tan45 * tanθ) + (tanθ * tan45)] / (1 - tan45 * tanθ) + [(tan45 * tanθ) - (tanθ * tan45)] / (1 + tan45 * tanθ)
Since tan45 = 1, we can further simplify:
[(tanθ + 1) / (1 - tanθ)] + [(tanθ - 1) / (1 + tanθ)]
Next, let's find the common denominator for the two fractions:
[(tanθ + 1)(1 + tanθ) + (tanθ - 1)(1 - tanθ)] / [(1 - tanθ)(1 + tanθ)]
Expanding the numerator:
[(tanθ + tanθ^2 + 1 + tanθ - 1 - tanθ + tanθ^2 -1) / (1 - tanθ * tanθ)]
Combining like terms:
[2tanθ + 2tanθ^2] / (1 - tanθ^2)
Since sec^2θ = 1 + tan^2θ, we can replace 1 - tanθ^2 with sec^2θ:
[2tanθ + 2tanθ^2] / sec^2θ
Finally, simplifying the expression:
2tanθ(sec^2θ) / sec^2θ
2tanθ
Therefore, we have proved that tan(45+θ) + tan(45-θ) = 2sec^2θ.
To prove the identity tan(45+θ) + tan(45-θ) = 2sec^2(θ), we can use the following steps:
Step 1: Start with the left-hand side of the equation and substitute the tangent angle addition formula:
tan(A+B) = (tan(A) + tan(B)) / (1 - tan(A)tan(B))
tan(45+θ) + tan(45-θ) = [(tan(45) + tan(θ))/(1 - tan(45)tan(θ))] + [(tan(45) - tan(θ))/(1 + tan(45)tan(θ))]
Step 2: Simplify the expression using the properties of tangents and the unit circle:
tan(45°) = 1, since the tangent of 45 degrees is 1
tan(θ)/tan(45) = sin(θ)/cos(θ) since tan(θ) = sin(θ)/cos(θ)
[(1 + tan(θ))/(1 - tan(θ))] + [(1 - tan(θ))/(1 + tan(θ))]
= [(1 + sin(θ)/cos(θ))/(1 - sin(θ)/cos(θ))] + [(1 - sin(θ)/cos(θ))/(1 + sin(θ)/cos(θ))]
Step 3: Combine the fractions by finding a common denominator:
= [(1 + sin(θ)/cos(θ))] x [(cos(θ)/cos(θ))/(cos(θ)/cos(θ))] + [(1 - sin(θ)/cos(θ))] x [(cos(θ)/cos(θ))/(cos(θ)/cos(θ))]
= [(cos(θ) + sin(θ))/(cos(θ) - sin(θ))] + [(cos(θ) - sin(θ))/(cos(θ) + sin(θ))]
Step 4: Add the two fractions together:
= [(cos(θ) + sin(θ))(cos(θ) + sin(θ))] + [(cos(θ) - sin(θ))(cos(θ) - sin(θ))] / [(cos(θ) - sin(θ))(cos(θ) + sin(θ))]
= [cos^2(θ) + 2sin(θ)cos(θ) + sin^2(θ) + cos^2(θ) - 2sin(θ)cos(θ) + sin^2(θ)] / [cos^2(θ) - sin^2(θ)]
= [2cos^2(θ) + 2sin^2(θ)] / [cos^2(θ) - sin^2(θ)]
Step 5: Use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1:
= [2(cos^2(θ) + sin^2(θ))] / [(cos^2(θ) - sin^2(θ))]
= [2(1)] / [(cos^2(θ) - sin^2(θ))]
= 2 / (cos^2(θ) - sin^2(θ))
= 2 / cos^2(θ) - 2 / sin^2(θ)
= 2sec^2(θ)
Thus, we have shown that tan(45+θ) + tan(45-θ) equals 2sec^2(θ).
(1+tan x)/(1-tan x) + (1-tan x)/(1+tan x)
= [(1+tan x)^2 + (1-tan x)^2]/(1-tan^2 x)
= (2 + 2tan^2 x)/(1-tan^2 x)
= 2(cos^2 x + sin^2 x)/(cos^2 x - sin^2 x)
= 2/cos 2x
= 2sec 2x