Prove that tan^2(alpha)-1+cos^2(alpha)sin^2(alpha)

Please Explain

I don't see any equal sign. What's to prove?

If sin A = 0.8and A lies between 90°&180° find value of tan1/2A

To prove the identity tan^2(alpha) - 1 + cos^2(alpha)sin^2(alpha), we need to manipulate and simplify the equation using trigonometric identities.

Let's start by expanding and simplifying the expression on the right side:

cos^2(alpha)sin^2(alpha) can be rewritten as (1 - sin^2(alpha))sin^2(alpha) using the identity cos^2(alpha) = 1 - sin^2(alpha).

Now we have: tan^2(alpha) - 1 + (1 - sin^2(alpha))sin^2(alpha)

Next, distribute the sin^2(alpha):

tan^2(alpha) - 1 + sin^2(alpha) - sin^4(alpha)

Combining the terms, we have:

tan^2(alpha) + sin^2(alpha) - sin^4(alpha) - 1

Now, let's take a closer look at the terms:
- tan^2(alpha) is equal to (sin^2(alpha))/(cos^2(alpha)).
- sin^2(alpha) is equal to 1 - cos^2(alpha).

Substituting these values into our expression, we get:

[(sin^2(alpha))/(cos^2(alpha))] + [1 - cos^2(alpha)] - sin^4(alpha) - 1

Next, let's get a common denominator for the first term:

[(sin^2(alpha))/(cos^2(alpha))] = [(sin^2(alpha))(cos^(-2)(alpha))]

Now, combine the terms:

[(sin^2(alpha))(cos^(-2)(alpha))] + [1 - cos^2(alpha)] - sin^4(alpha) - 1

Using a common denominator, we can rewrite the terms:

[(sin^2(alpha))(cos^(-2)(alpha))] = [(sin^2(alpha))(1/cos^2(alpha))]

Now, we can multiply the numerators and denominators:

[(sin^2(alpha))(1/cos^2(alpha))] = [(sin^2(alpha))/(cos^2(alpha))] = tan^2(alpha)

The expression simplifies to:

tan^2(alpha) + 1 - cos^2(alpha) - sin^4(alpha) - 1

Combine like terms:

tan^2(alpha) - cos^2(alpha) - sin^4(alpha)

Now, let's use another trigonometric identity:

cos^2(alpha) = 1 - sin^2(alpha)

Substituting this into our expression:

tan^2(alpha) - (1 - sin^2(alpha)) - sin^4(alpha)

Distribute the negative sign:

tan^2(alpha) - 1 + sin^2(alpha) - sin^4(alpha)

The expression matches the original equation:

tan^2(alpha) - 1 + cos^2(alpha)sin^2(alpha)

Therefore, we have proved the given identity.