Pre Calc- Please Help.?

Question

Please Help me, I'm very confused, and these don't seem to be coming out right. My textbook and notes aren't helping.

1- [@ = theta] Solve for ALL numeric values of @:

tan @+8/5 + tan@-1/tan @ = 1

2- Prove the identities:

(a.) tan x + cot x/ (cos x) (sin x) = (sec^2 x)(csc^2 x)

(b.) sin 2x= 2 tan x/1+ tan^2 x

3- The angle of elevation from the ground to the top of a toper is 40 degrees. A person then walks 300 meters closer to the tower and the angle of elevation is now 60 degrees. (a) What is the height of the tower (in meters)?
(b) What is the distance from point A to point B the distance the person walks (in meters)?


*Please help explain these to me step by step, I have to understand these concepts for an upcoming exam. I would really appreciate the help. Thanks!

Answer 1

1. (My apologies if I'm not interpreting this question correctly - if some other order of operations should be used, please use brackets where required.)
tan θ + 8/5 + tan θ - 1/tan θ = 1
<=> 2 tan θ + 3/5 - 1/tan θ = 0
<=> 10 tan^2 θ + 3 tan θ - 1 = 0
<=> (5 tan θ - 1) (2 tan θ + 1) = 0
<=> tan θ = 1/5 or -1/2
<=> theta; = 0.1974 + k&pi, -0.4636 + kπ for k any integer.
(answers are in radians)

2.(a) LHS = (tan x + cot x) / (cos x sin x)
= (sin x / cos x + cos x / sin x) / (cos x sin x)
= (sin^2 x / (cos x sin x) + cos^2 x / (cos x sin x)) / (cos x sin x)
= ((sin^2 x + cos^2 x) / (cos x sin x)) / (cos x sin x)
= (1 / (cos x sin x)) / (cos x sin x)
= 1 / (cos^x sin^2 x)
= sec^2 x csc^2 x
= RHS.

(b) RHS = 2 tan x / (1 + tan^2 x)
= 2 tan x / sec^2 x
= 2 tan x (cos^2 x)
= 2 (sin x / cos x) (cos^2 x)
= 2 sin x cos x
= sin 2x
= LHS.

3). Suppose they were initially x metres from the tower and the height of the tower is h metres.
Then we know:
tan 40° = h / x
tan 60° = h / (x-300)
So h = x tan 40° = 0.8391 x (4 s.f.)\
h = (x-300) tan 60° = √3 x - 300√3
So √3 x - 300√3 = 0.8391 x
<=> (√3 - 0.8391) x = 300√3
<=> 0.8930 x = 519.6 (4 s.f.)
<=> x = 581.9
and h = 0.8391 (581.9) = 488.3
So the tower is 488 m high. (3 s.f.)
I'm not sure what (b) is supposed to be asking - we are given that the distance the person walks from point A to point B is 300m. If you want to know how far A and B are from the tower, the answer is 582 m and 282 m respectively.

Answer 2

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