​Pre calc help! I did some questions but I don’t understand how to do the others!! :(

tan  ( arccos (2/3) )

We are looking for the tangent  of an angle whose cosine is  2/3

So..we know  x / r   and we are looking for y/ x...so we have

sqrt  ( 3^2 - 2^2)  =  sqrt 5   =  y

So....y / x  =     sqrt(5) / 2

cos ( arcsin (-3/5) )

We are looking for the cosine .... (x / r)..... of an angle whose sin  is  -3/5  =  y / r

So  we need to find x

sqrt [ 5^2  - (-3)^2  ] = sqrt [ 25 - 9 ] =  4  = x

This angle is in the 4th quadrant....the cosine is positive here

So...this evaluates to  x / r  =   4/5

sec (arcos (x) )

The cosine of this angle =  x / 1

So..the sec  is the reciprocal,  1 / x

cot  ( arccos (3x) )

The cosine of this angle  =   [3x] / 1    =  3x

We are looking for the  cot =    cos / sine

So...the sine  =

sqrt  [ 1 - [3x]^2 ]  =  sqrt [ 1 - 9x^2 ]

And the  cotangent  is

[ 3x]  / sqrt [ 1 - 9x^2 ] 

cos  [ 2 arcsin ( x) ]

Let  arcsin (x)  be θ

So...using a trig identity....  we have

cos ( 2θ)  =    1  - 2sin^2(θ)  =  1 - 2 [ sin (θ) ]^2

The sin of θ  =  x / 1  =  x

So...

cos (2θ)  =  1 - 2sin^2 ( θ)  =  1 -  2 [ sin (θ)]^2  =   1  - 2 [x]^2  =  1  - 2x^2