If A + B = 90 ° then, show that, 1 + cot A cot B = cosec 2 A .

Given, A + B = 90 ° i.e., B = 90 ° - A L . H . S = 1 + cot A cot B = 1 + cot A cot ( 90 ° - A ) We know that, cot ( 90 ° - θ ) = tanθ = 1 + cot A tan A = 1 + cot 2 A ( ∴ cosec 2 A - cot 2 A = 1 ) ⇒ cosec 2 A = 1 + cot 2 A = cosec 2 A ∴ L . H . S = R . H . S . Hence, proved.

If A and B are two complementary angles, then prove that, cos B cos A ( 1 + sin B ) = cos B + cot B

Given, A + B = 90 ° ⇒ A = 90 ° - B ⇒ A = π 2 - B L.H.S = cos B cos A ( 1 + sin B ) = cos B cos A + cos B sin B cos A = cos B cos ( π / 2 - B ) + cos B sin B cos ( π / 2 - B ) = cos B sin B + cos B sin B sin B = cot B + cos B = R.H.S Therefore, L . H . S = R . H . S

If A and B are complementary angles, then prove that, 1 + tan A tan B = tan 2 A sec 2 B

Given, A + B = 90 ° ⇒ B = 90 ° - A L.H.S = 1 + tan A tan B = 1 + tan A tan ( 90 ° - A ) = 1 + tan A cot A = 1 + tan A · tan A = 1 + tan 2 A = sec 2 A Now, R.H.S = tan 2 A s e c 2 B = tan 2 A · sec 2 90 ° - A = tan 2 A · cosec 2 A = sin 2 A cos 2 A × 1 sin 2 A = 1 cos 2 A = s e c 2 A = R.H.S Therefore, L . H . S = R . H . S