If there is a plus in between, that would be equal to 1, as a result of the Pythagorean Theorem. Otherwise, you can convert this into other forms with some of the trigonometric identities for multiplication, but you won't really get it into a simpler form.
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
It's 1/2 of sin(2 theta) .
tan θ = sin θ / cos θ sec θ = 1 / cos θ sin ² θ + cos² θ = 1 → sin² θ - 1 = - cos² θ → tan² θ - sec² θ = (sin θ / cos θ)² - (1 / cos θ)² = sin² θ / cos² θ - 1 / cos² θ = (sin² θ - 1) / cos² θ = - cos² θ / cos² θ = -1
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
Your question is insufficiently precise, but I'll try to answer anyway. "Sine squared theta" usually means "the value of the sine of theta, quantity squared". "Sine theta squared" usually means "the value of the sine of the quantity theta*theta". The two are not at all the same.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
1
Cos theta squared
COS squared Theta + SIN squared Theta = 1; where Theta is the angles measurement in degrees.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
Tan^2