Prove that:
(tan²Θ)x(sin²Θ) = (tan²Θ)-(sin²Θ)
show work
LS = (sin^2 Ø)/cos^2 Ø) * sin^2 Ø
= sin^4 Ø/cos^2 Ø
RS = sin^2 Ø / cos^2 Ø - sin^2 Ø
= (sin^2 Ø - sin^2 Ø(cos^2 Ø) )/cos^2 Ø
= sin^2 Ø( 1 - cos^2 Ø)/cos^2 Ø
= sin^2 Ø (sin^2 Ø) /cos^ Ø
= sin^4 Ø / cos^2 Ø
= LS
Thank You!!!
True/False.
sec Θ ° > tan Θ ° for every angle Θ.
Please explain why?
To prove the given equation, we need to simplify both sides of the equation and show that they are equal.
First, let's simplify the left side of the equation:
(tan²Θ) x (sin²Θ)
Using the trigonometric identity tan²Θ = sin²Θ / cos²Θ, we can substitute this into the expression:
(sin²Θ / cos²Θ) x (sin²Θ)
Now, let's simplify further by multiplying the numerators together and the denominators together:
(sin²Θ x sin²Θ) / (cos²Θ)
Next, let's simplify the numerator:
sin²Θ x sin²Θ = (sinΘ x sinΘ) x (sinΘ x sinΘ) = sin⁴Θ
Now, substitute sin⁴Θ back into the expression:
sin⁴Θ / (cos²Θ)
Now, let's simplify the right side of the equation:
(tan²Θ) - (sin²Θ)
Again, we can substitute tan²Θ = sin²Θ / cos²Θ:
(sin²Θ / cos²Θ) - (sin²Θ)
Now, let's find a common denominator and combine the terms:
(sin²Θ - sin²Θ x cos²Θ) / cos²Θ
Simplifying further:
(sin²Θ - sin⁴Θ) / cos²Θ
Therefore, the left side of the equation simplifies to sin⁴Θ / (cos²Θ), and the right side simplifies to (sin²Θ - sin⁴Θ) / cos²Θ.
To show that these two sides are equal, we need to set them equal to each other and solve for Θ.
(sin⁴Θ / cos²Θ) = (sin²Θ - sin⁴Θ) / cos²Θ
Multiplying both sides of the equation by cos²Θ:
sin⁴Θ = sin²Θ - sin⁴Θ
Adding sin⁴Θ to both sides of the equation:
2sin⁴Θ = sin²Θ
Dividing both sides of the equation by sin²Θ:
2sin²Θ = 1
Dividing both sides of the equation by 2:
sin²Θ = 1/2
Using the trigonometric identity sin²Θ + cos²Θ = 1, we can substitute this into the equation:
1/2 + cos²Θ = 1
Subtracting 1/2 from both sides:
cos²Θ = 1 - 1/2
cos²Θ = 1/2
Taking the square root of both sides:
cosΘ = √(1/2)
Since we know that sin²Θ = 1/2 and cosΘ = √(1/2), we can use these values to find the angle Θ.
By taking the inverse sine (sin⁻¹) of √(1/2), we find that Θ = π/4 or 45 degrees.
Therefore, the equation (tan²Θ)x(sin²Θ) = (tan²Θ)-(sin²Θ) is true for Θ = π/4 or 45 degrees.