Question

Prove that sin²A tan²B-cos²Asin²B=cos²B-cos²A​

Answer 1

1. (sinx)^2 +(cosx)^2=1

Let this be equation 1

2. cos2x = 2[(cosx)^2] -1

Let this be equation 2

3. cos2x = 1 - 2[(sinx)^2]

Let this be equation 3

where x=A or x =B

It is given that

(cosA)^2 - (sinA)^2 = (tanB)^2

Let this be equation 4

From equation 1 we get

(cosA)^2 + (sinA)^2 = 1

Adding the above equation to equation 4, we get

2[(cosA)^2] = 1 + (tanB)^2

We know that tanB= sinB/cosB

Hence,

2[(cosA)^2] -1 = (sinB)^2/(cosB)^2

Using equation 2, we get

cos2A = (sinB)^2/(cosB)^2

Let this be equation 5

Using equation 1 in 5, we get

cos2A = (sinB)^2/[1-(sinB)^2]

Simplifying,

[1-(sinB)^2]cos2A = (sinB)^2

⇒ cos2A - [(sinB)^2]cos2A = (sinB)^2

⇒ cos2A = [(sinB)^2](1 + cos2A)

⇒ (sinB)^2 = cos2A/(1 + cos2A)

Now instead of changing tanB(in equation 5) into terms of sinB we change it into terms of cosB

Hence, using equation 1 to simplify equation 5, we get

cos2A = [1-(cosB)^2]/(cosB)^2

⇒ [(cosB)^2]cos2A = 1 - (cosB)^2

⇒ 1 = [(cosB)^2](1 + cos2A)

⇒ (cosB)^2 = 1/[1 + cos2A]

Hence ,

(cosB)^2 - (sinB)^2 = 1/[1 + cos2A] - cos2A/(1 + cos2A)

=[1 - cos2A]/[1+cos2A]

Using equations 2 and ,3 we get

(cosB)^2 - (sinB)^2 = [2(sinA)^2]/[2(cosA)^2]

=(tanA)^2