Assume cos(t) = 4 where sin(-t) = csc(t + 2) = - cot(t)= 3π < t < 2π. Compute the following: 2

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

A: Here  given as Use some trigonometric formula as

Q: sec(t) – cos(t) - Simplify to a single trig function. sin(t)

A: Using trigonometric formulas

Q: Verify that (sin(0) csc(0))+ (cot(0) tan(0)) = 2

A: We have to verify,

Q: sin(2w)dw =

A:

Q: Find the function if sin t T+1 1 tan 2x + 1

A:

Q: ] 1 1-x² dx = tan¹(x)

A:

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A: Given problem is sin(t)-cos(t)=cot(t)

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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: | cos (nt)sin(at) dt Find

A: This integral can be solved by the method of substitution.

Q: If csc(x)=8, for 90 degrees&lt;(x)&lt;180 degrees, then what is sine(x/20, cos(x/2), and tan(x/2)?

A:

Q: Show that tan-1 ( 1/v ) = π/2 - tan-1 v, if v &gt; 0.

A: To prove tan-11v=π2-tan-1v, let  x=tan-1v then tan x=v1tan x=1v Since  tanx=1cotxcotx=1v ⋯ (1)

Q: 大 1-cos X P-V= 2

A:

Q: Write f(x) = -2 sin(x) + 4 cos(x) using a single sine function.

A:

Q: Graph the epitrochoid with equations x  =11 =cos t - 4 cos(11t/2), y = 11 sin t - 4 sin(11t /2)

A:

Q: find sec(A) and sin(0).

A:

Q: S cos(e) In (sin(0))d0

A: We take the help of integration by parts in solving this problem

Q: where does the function 1/sin(x) not exist? what happens if we try to calculate any of these values?

A: We have to find the values where function values doesn't exist.

Q: tan(cos-1(x/4))

A: We will consider a right triangle where adjacent= x and hypotenuse= 4 In this triangle