Question

11. If A+B = 450°, then prove that :-
(i) (1+tanA) (1 + tanB)=2
(ii) (CostA-1) (CotB-1) = 2.​

Answer 1

Answer:

1. A + B = 45'

Apply tan on both sides

tan(A + B) = tan 45'

$\frac{ \tan(a) + \tan(b) }{1 - \tan(a) \tan(b) } = 1$

Tan A + tanB = 1 - Tan A Tan B

Tan A + tanB + Tan A Tan B = 1

Add one on both sides

Tan A + tanB + Tan A Tan B + 1 = 1 +1

Tan A + Tan A Tan B + tan B + 1 = 2

Tan A (1 + tan B) + (1 + tan B) =2

(Tan A + 1) + ( 1 + tan B )= 2

2. A + B = 45'

Apply cot on both sides

Cot (A + B) = cot 45'

$\frac{ \cot(a) \cot(b) \: - 1 }{ \cot(b) + \cot(a) } = 1$

Cot A cot B - 1 = cot B + cot A

Cot A cot B - cot B - cot A = 1

Add one on both sides

Cot A cot B - cot B - cot A + 1 = 1 + 1

Cot A cot B - cot B - cot A + 1 = 2

Cot B ( cot A - 1) - 1 (cot A + 1) = 2

(Cot A - 1 ) (Cot B - 1) = 2

Hence proved