Hiya, I'm stuck on the question sinx^2(cotx+1)^2 = cosx^2(tanx+1)^2 essentially we need to prove the identity. Whenever "^2" is used it means squared. Such as 2^2 = 2 squared

Use the reduction formulas to reduce sin^4 θ to only linear powers of cosine (cos(θ) = ”linear”, (cos^(any power > 1)θ) = not ”linear”)

For this problem, fill in each blank with the correct answer. Syntax example: to enter r? sin (), type x^2sin(x/2) (do not use any space or * for multiplication and use parentheses for the input/argument of the sine function). To enter x? (- sin(:), type x^2(- sin(x/2)) (use parentheses when multiplying by a negative and do not use any space or * for multiplication) or just simplify to -x^2sin(x/2). It is always best to simplify your answers as much as possible. (For instance, if the answer is 2/(1/2), simplify to 4, if the answer is 2(-3) simplify to -6, if the answer is 2x(-x^2) simplify to -2x^3, etc.) Note: DO NOTI include the "dx" part in your answers below! It's already there. Evaluate the indefinite integralf x? sin ()da by following the steps outlined below. We will use the method of integration by parts which is derived from the for differentiation. We first apply integration by parts with , dv = dx du = dx, v = to get: Ja? sin(플) dz= +4/zcos (특) da. Next, we apply integration…

differentiate y = sin^3(tet^2) and simplify

Please prove identity by transforming the left side into the right side: (1+secx)(1-cosx) = (tan^2 x)( cosx) (please note tan^2 means tan squared)

theta is not -tan^-1(5/2)

Write the trigonometric expression as an algebraic expression in u. sec (cos^-1 u)

please show all steps of how to verify identity  (1-cosx)/(sinx)   +   (sinx)/(1-cosx)  = 2cscx

In my calculus 3 class, my professr is talked about the derivation of an equation. However, I feel confused about how can |a|^2*|b|^2*cos^2 equal to (1-cos^2). The topic is about corss product. the above equation is on the fifth row of the submitted image.

Ok, I see how sinx derives to cosx using that formula sin^2x + cos^2x = 1, thx it helps