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.