Question

If A + B = 45theta, Then show that (1 + tan A) (1 + tan B) = 2

Answer 1

Step-by-step explanation:

Since A + B = 45°

∴ tan (A + B) = tan 45° = 1

or, (tan A + tan B) / (1 - tan A tan B)  = 1

This gives, tan A + tan B = 1 – tan A tan B

So, tan A + tan B + tan A tan B = 1

tan A  (1 + tan B) + tan B + 1 = 2

Hence, (1 + tan B) (1 + tan A) = 2

Let us check:

(sin A cos B + cos A sin B) (sin A cos B – cos A sin B)

= sin^2A cos^2B – cos^2A sin^2B

= sin^2A (1 – sin^2B) – (1 – sin^2A) (sin^2B)

= sin^2A – sin^2A sin^2B – sin^2B + sin^2A sin^2B

= sin^2A – sin^2B

= RHS

Answer 2

Answer:

hope it will thanks for asking

Step-by-step explanation:

theta 3 by 2