'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
The identity for tan(theta) is sin(theta)/cos(theta).
The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant
The equation cannot be proved because of the scattered parts.
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
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
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).
It's 1/2 of sin(2 theta) .
The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
because sin(2x) = 2sin(x)cos(x)
The equation cannot be proved because of the scattered parts.
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
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
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.