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Prove tan^2 theta - tan^2 phi = (sin^2 theta- sin^2 phi) /cos^2 theta cos^2phi

User Shane Lu
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Answer:

tan^2 theta - tan^2 phi = (sin^2 theta- sin^2 phi) /(cos^2 theta cos^2phi) (identity has been verified)

Explanation:

Verify the following identity:

tan(θ)^2 - tan(ϕ)^2 = (sin(θ)^2 - sin(ϕ)^2)/(cos(θ)^2 cos(ϕ)^2)

Hint: | Eliminate the denominator on the right hand side.

Multiply both sides by cos(θ)^2 cos(ϕ)^2:

cos(θ)^2 cos(ϕ)^2 (tan(θ)^2 - tan(ϕ)^2) = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Express the left hand side in terms of sine and cosine.

Write tangent as sine/cosine:

cos(θ)^2 cos(ϕ)^2 ((sin(θ)/cos(θ))^2 - (sin(ϕ)/cos(ϕ))^2) = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Simplify the left hand side.

cos(θ)^2 cos(ϕ)^2 ((sin(θ)/cos(θ))^2 - (sin(ϕ)/cos(ϕ))^2) = cos(θ)^2 cos(ϕ)^2 ((sin(θ)^2)/(cos(θ)^2) - (sin(ϕ)^2)/(cos(ϕ)^2)):

cos(θ)^2 cos(ϕ)^2 (sin(θ)^2/cos(θ)^2 - sin(ϕ)^2/cos(ϕ)^2) = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Put the fractions in sin(θ)^2/cos(θ)^2 - sin(ϕ)^2/cos(ϕ)^2 over a common denominator.

Put sin(θ)^2/cos(θ)^2 - sin(ϕ)^2/cos(ϕ)^2 over the common denominator cos(θ)^2 cos(ϕ)^2: sin(θ)^2/cos(θ)^2 - sin(ϕ)^2/cos(ϕ)^2 = (cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2)/(cos(θ)^2 cos(ϕ)^2):

cos(θ)^2 cos(ϕ)^2 (cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2)/(cos(θ)^2 cos(ϕ)^2) = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Cancel down (cos(θ)^2 cos(ϕ)^2 (cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2))/(cos(θ)^2 cos(ϕ)^2).

Cancel cos(θ)^2 cos(ϕ)^2 from the numerator and denominator. (cos(θ)^2 cos(ϕ)^2 (cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2))/(cos(θ)^2 cos(ϕ)^2) = (cos(θ)^2 cos(ϕ)^2 (cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2))/(cos(θ)^2 cos(ϕ)^2) = cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2:

cos(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2 = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Express cos(ϕ)^2 in terms of sine via the Pythagorean identity.

cos(ϕ)^2 = 1 - sin(ϕ)^2:

1 - sin(ϕ)^2 sin(θ)^2 - cos(θ)^2 sin(ϕ)^2 = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Express cos(θ)^2 in terms of sine via the Pythagorean identity.

cos(θ)^2 = 1 - sin(θ)^2:

sin(θ)^2 (1 - sin(ϕ)^2) - 1 - sin(θ)^2 sin(ϕ)^2 = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Expand (1 - sin(ϕ)^2) sin(θ)^2.

(1 - sin(ϕ)^2) sin(θ)^2 = sin(θ)^2 - sin(θ)^2 sin(ϕ)^2:

sin(θ)^2 - sin(θ)^2 sin(ϕ)^2 - sin(ϕ)^2 (1 - sin(θ)^2) = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Expand -(1 - sin(θ)^2) sin(ϕ)^2.

-(1 - sin(θ)^2) sin(ϕ)^2 = sin(θ)^2 sin(ϕ)^2 - sin(ϕ)^2:

sin(θ)^2 - sin(θ)^2 sin(ϕ)^2 + sin(θ)^2 sin(ϕ)^2 - sin(ϕ)^2 = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Evaluate sin(θ)^2 - sin(θ)^2 sin(ϕ)^2 - sin(ϕ)^2 + sin(θ)^2 sin(ϕ)^2.

sin(θ)^2 - sin(θ)^2 sin(ϕ)^2 - sin(ϕ)^2 + sin(θ)^2 sin(ϕ)^2 = sin(θ)^2 - sin(ϕ)^2:

sin(θ)^2 - sin(ϕ)^2 = ^?sin(θ)^2 - sin(ϕ)^2

Hint: | Come to a conclusion.

The left hand side and right hand side are identical:

Answer: (identity has been verified)

User Harshit Agarwal
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