# Trigonometry help .....?

1. If tan A = n tan B , sin A = m sin B, prove that cos ^ 2 A = (m ^ 2 - 1) / (n ^ 2 - 1)
2. If cos ^ 2 a - sin ^ 2 a = tan ^ 2 b, prove that tan ^ 2 a = cos ^ 2 b - sin ^ 2 b
3. If cosec a - sin a = m and sec a - cos a = n, prove that [ { (m ^ 2) * n} ^ 2/3 ] + [ { m * ( n ^ 2 ) } ^ 2/3 ] = 1

Tan A = n Tan B

Divide each side by ' Tan B '

n = Tan A / Tan B

Quotient Identities is TanA = sinA/cosA

n = [sinA/cosA] / [sinB/cosB]

Then n = sinAcosB / cosAsinB

And sinA = m sinB

Divide each side by ' sin B '

m = sinA / sinB

Now take  (m2-1) / (n2-1)

= [(sinA / sinB)2 -1] / [(sinAcosB / cosAsinB)2-1]

Rewrite the expression with common denominator.

= [(sin2A - sin2B) / sin2B] / [(sin2Acos2B - cos2Asin2B) / (cos2Asin2B)]

Simplify (sin2A - sin2B)(cos2Asin2B) / (sin2B)(sin2Acos2B - cos2Asin2B)

Pythagorean Identities: sin2θ = 1 - cos2θ

= [(1 - cos2A) - (1 - cos2B)]cos2A(1 - cos2B) / (1 - cos2B)[(1 - cos2A)( cos2B) - cos2A(1 - cos2B)

= [ - cos2A  + cos2B)]cos2A(1 - cos2B) / (1 - cos2B)[(cos2B - cos2A cos2B - cos2A + cos2Acos2B

=  (cos2B - cos2A)cos2A(1 - cos2B) / (1 - cos2B) [cos2B - cos2A]

= cos2A (cos2B - cos2A)(1 - cos2B) / (1 - cos2B) [cos2B - cos2A]

= cos2A

There fore

(m2 - 1) / (n2 - 1) = cos2A

cosec a-sin a = m and sec a-cos a = n,

Recall reciprocal identities is coses A = 1/sinA and secA = 1/cosA

cosec a - sin a = m

(1/sin a) - sin a = m

Rewrite the expression with common denominator.

(1 - sin2 a)/sin a = m

Pythagorean Identities is 1 - sin2 θ = cos2θ

cos2a/sin a = m

m = cos2a/sin a

And sec a-cos a = n

Similarly n = sin2a/cos a

Now take [{m2×n}2/3] + [{m×(n2)}2/3]

m2×n  = [cos2a/sin a]2×[sin2a/cos a]

= (cos4a)(sin2a)/(sin2a)(cos a)

Simplify m2×n = cos3a

{m2×n}2/3 = (cos3a)2/3 = cos2a

m×n2 = (cos2a/sin a)×(sin2a/cos a)2

= (cos2a)(sin4a)/(sin a)(cos2a)

Simplify m×n2=sin3a

{m×n2}2/3 = {sin3a}2/3 = sin2a

[{ m2×n}2/3 + { m×n2}2/3] = cos2a + sin2a

So, [{ m2×n}2/3 + { m×n2}2/3] = cos2a + sin2a = 1

There fore [{ m2×n}2/3 + { m×n2}2/3] = 1