Question

Cos²A-sin²A= tan B. Prove that cos²B-sin²B=tan²A

Answer 1

Step-by-step explanation:

Given : cos²A - sin²A = tan²B

We know that,

cos²A + sin²A = 1

First we add these equations, we get

→ 2cos²A = 1 + tan²B

→ cos²A = (1 + tan²B)/2

→ cos²A = sec²B/2

{ sec²∅ = 1 + tan²∅ }

Second we subtract these equations, we get

→ 2sin²A = 1 - tan²B

→ sin²A = (1 - tan²B)/2 ...(ii)

We know that,

sin²A/cos²A = tan²A

→ [ (1 - tan²B)/2 ] / [ sec²B/2 ] = tan²A

[ using (i) & (ii) ]

→ (1 - tan²B) / sec²B = tan²A

→ (1/sec²B) - (tan²B/sec²B) = tan²A

→ (1/sec²B) - [ (sin²B/cos²B) / (1/cos²B) ] = tan²A

→ cos²B - sin²B = tan²A

Hence, proved !!