10. Solve for t on the interval [0, 2π) (a) 2 cos² t-cost-1=0 (b) 2 sin²t-3 sint + 1 = 0 taol al (c) tan² t-tant=0up att, not (d) sec2t-3 sect+2=0

Q: Q4// Find the value of csc(sec-1 2) + cos(tan=²(-V3)).

A: Here  given as Use some trigonometric formula as

Q: QI) Show that: (a) (1+x)° =1+6x (b) (cosx) = –sinx dx

A: 1.  a  1+x6≅1+6xb  ddxcos x=-sin x

Q: 6. sec (sin-1 x)

A: Interconversion

Q: (6) [²/2 sin² (2) cos2 (x).dx = A) 플 B) 플 C) 미플

A: We need to make it proper to solve this integration.

Q: 4 sec(8) = find cot(0) 2

A: Use trigonometric rations

Q: S(sinx – cosx)²dx COoSx

A: Given, ∫sinx-cosx2dx  We have to find the integral.

Q: csch (2 sin2x) dx 1. V1-4x² r(²-1)cos[cosh (In r)]dr OS 4.

A:

Q: Find the first and second derivatives of r = [3 cos 2t, 3 sin 2t , 4t].

A:

Q: sinx

A:

Q: C- Find f' of the following: 1) √2x³-2² + In ² 2) 3xe³r+1 3) sin ¹(³).

A:

Q: 7/4 (sec² x – sin x)dx

A: 1. ∫sec2x dx=tanx 2. ∫sinx dx=-cosx 3.∫(u±v)dx=∫u dx±∫v dx 4. ∫abf(x) dx=F(b)-F(a), where ∫f(x)…

Q: T/2 cosx dx (4 +2 sin x)3

A:

Q: a) Use suitable sum and difference identities for hyperbolic functions to show that sinh(5x)…

A:

Q: 2 2. (sin cos x) = sin 2x +1 2х +

A: The given identity

Q: Find y' , for y=Sin-1(Cos√x)+Cos-1(Sin√x)

A: The solution for the above question is as shown below.

Q: Find cos(u+v) sinu= -4/5 with 3pi/2 <u<2pi and cosv= -5/13 with pi/2 <v<pi

A: u is in the fourth quadrant. There cos will be positive.

Q: Find , if z= cos(3x² + 7y°) O 2x sin(3a? + 7y³) sin(3x² + 7y³) -9x cos(3x2 + 7y°) -6x sin(3x? + 7y°)

A:

Q: Solve the Bernoulli's equation given below: dy 2 dx -y cot2x = y' csc2x O A. [(y )^(- 2)] sin 2x = -…

A: Use Bernoulli theorem to solve

Q: Solve for x Cos(2/3x+15)=sin(1/4x+60)

A:

Q: If x*y" + xy' + (x? - y= 0 x >0 ,V1=x 2 sinx 3D0 and y =y,v. Which of the following equations…

A:

Q: 2. Find exact value of (a) sin-(-1) (b) cos-11 (c) tan-1 /3 (d) sin(sin¬' 0.8) (e) tan-'(tan ) (f)…

A: (a) Consider the given expression: sin-1-1 Sine gives negative one as it value at -π2. Therefore,…