plz plz help trigonometry?

Question

show work plz

1. Express cot²x-1/csc²x in terms of sin x.


2. Reduce (csc²x-sec²x) to an expression containing only tan x.


3. Verify the following identities.
a. (sin B + cos B) (sin B – cos B) = 2 sin² B -1
b.(1 – cos² y + sin² y)² +4 sin² y cos² y=4 sin² y
c. tan² 0 sec² 0- sec² 0+1 = tan 0
d. sin 0/csc 0 + cos 0/ sec 0 =1
e. sin x tan ² x cot ^3 x=cos x
f. sec² x/sec ² x-1 =csc² x

Answer 1

Answer 1:
cot²x = (cot x)(cot x)
= (1/tan x)(1/tan x)
= 1/tan²x
= 1/(sin²x/cos²x)
= (cos²x/sin²x)
= (1 - sin²x/sin²x)
csc²x = 1/sin²x
cot²x-1/csc²x = (1 - sin²x/sin²x)/(1/sin²x)
= (1 - sin²x/sin²x) * sin²x
= 1 - sin²x

Use similar identities to complete the second one.
I'm no good at identity verification. I'm only in Grade 9 in India. They teach us Trig only in Class 10. But still, I managed the first one!

Answer 2

ANS2:
cosec^2 x - sec^2 x
=(cot² x+1)-(tan² x+1)
=1/tan² x - tan² x

ANS3:
a) sin² B - cos² B
= sin² B - (1 - sin² B)
=2sin² B - 1
b) (1 – cos² y + sin² y)² +4 sin² y cos² y
= (1 - 2cos² y + 2sin² y - 2sin² y cos² y + cos^4 y + sin^4 y) +4 sin² y cos² y
= 1 - 2cos² y + 2sin² y + 2sin² y cos² y + cos^4 y + sin^4 y
= 1 - 2cos² y + 2sin² y + (sin² y + cos²y)²
= 2 - 2cos² y + 2sin² y
= 4sin² y
c) tan 0 = 0
sec 0 = 1
therefore tan² 0 sec² 0- sec² 0+1
d)sin 0/cosec0 + cos 0/sec0
=sin 0/1/sin 0 + cos 0/1/cos 0
=sin² 0 + cos² 0
=1
= 0 - 1 + 1 = 0 = tan 0
e)sin x tan ² x cot ^3 x
= sin x cot x
= sin x (cos x/sin x)
= cos x
f)sec² x/(sec ² x-1)
= 1/cos² xtan² x
= 1/sin² x
= cosec² x

Answer 3

4. locate cos (x+y) whilst the two x and y are acute angles and sin x = 3/5 and sin y= 5/13. Use the courting sin² ? + cos² ? = a million and remedy for cos ?, given sin ?. If sin x = 3/5 then cos x = 4/5. If sin y = 5/13 then cos y = 12/13. Now word the formula for attitude addition for cosine. cos(x + y) = (cos x)(cos y) - (sin x)(sin y) = (4/5)(12/13) - (3/5)(5/13) = (40 8 - 15)/(5*13) = 33/sixty 5