if tan27=p with the aid of a diagram express sin 63 , cos(-126),cos18 in terms of p without a calculator.?

tan27° = p

Sketch a right-angled triangle with angle A = 27°.
For tan27° = p, let p be the opposite side and 1 be the adjacent side.

By Pythagorean theorem :
The hypotenuse =√(p² + 1)

Hence, sin27° = (opposite side)/hypotenuse = p/√(p² + 1)
and cos27° = (adjacent side)/hypotenuse = 1/√(p² + 1)

sin63°
= cos(90° - 63°)
= cos27°
= 1/√(p² + 1)

cos(-126°)
= cos126°
= cos(180° - 54°)
= - cos(54°)
= - cos(2×27°)
= - (cos²27° - sin²27°)
= - {[1/√(p² + 1)]² - [p/√(p² + 1)]²}
= [p²/(p² + 1)] - [1/(p² + 1)]
= (p² - 1)/(p² + 1)

cos126°
= (p² - 1)/(p² + 1)

sin126°
= √{1 - cos²126°}
= √{1 - [(p² - 1)/(p² + 1)]²}
= √{[(p² + 1)/(p² + 1)]² - [(p² - 1)/(p² + 1)]²}
= √{[(p² + 1)² - (p² - 1)]/(p² + 1)²}
= √{[p⁴ + 2p² + 1 - p⁴ + 2p² - 1]/(p² + 1)²}
= 2p/(p² + 1)

cos18°
= cos(270° - 252°)
= - sin(252°)
= - sin(2×126°)
= - 2 sin126° cos126°
= - 2 [2p/(p² + 1)] [(p² - 1)/(p² + 1)]
= -4p(p² - 1)/(p² + 1)²
= p(1 - p²)/(p² + 1)²

tan = opp/adj (in a right triangle)
opp = p
adj = 1
hyp = sqrt(1+p^2) --- Pythogorean theorem

sin(63) = opp/hyp = 1/sqrt(1+p^2)

https://gyazo.com/b3f0af52a92fb7119ad2479a7ce38ee3
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tan(45) = 1
tan(27+18) = 1
(tan(27)+tan(18))/(1-tan(27)tan(18)) = 1
(p+tan(18)) /(1-tan(18)p ) = 1
p+tan(18) = 1- ptan(18)
tan(18) + ptan(18) = 1-p
tan(18)(1+p) = 1-p

tan(18) = (1-p) /(1+p) = opp/ adj
opp = 1-p
adj = 1+p
hyp = sqrt( (1-p)^2+(1+p)^2) = sqrt( 1-2p+p^2+1+2p+p^2) = sqrt(2+2p^2)

cos(18) = adj/hyp = (1+p) / sqrt(2+2p^2)

tan (27+63)=tan 90 = infinity, so invert.
1/ (tan (27+63))= cotan 90= 0
equivalent to tan (a+b)= tan a + tan b / (1-tan a tan b)

so we have
1- tan a tan b/ (tan a + tan b) = 0, let a= 27, b=63

1- tan 27 tan 63/ (doesn't matter the top must equal zero) =

1-ptan 63=0, so tan 63= 1/p

now tan a= sin a/cos a, and since cos a= (1- sin ^2 a) ^1/2

so
tan a = sin a / (1- sin ^2 a) ^1/2, square both sides

tan ^2 a = sin ^2 a / (1- sin ^2 a) tan 63= 1/p

so
1= p^2 sin ^a / (1- sin ^2 a)

1- sin ^2 a= p^2sin^2 a
1= sin^2 a(p^2+1)

sin a = ( 1/(p^2+1))^1/2 = sin 63

thank you every one for your help . I made the stupid mistake of not noticing that tan 27= p/1 . I appreciate all of your responses .