Question

If A+B=45 , then prove that (1+tanA) (1+tanB)=2​

Answer 1

Answer:

Given A+B=45

{Take tan on both the sides }

tan(A+B) = tan45

tanA+tanB/1- tanA tanB = 1

tanA+tanB=1-tanA.tanB

tanA+tanB+tanA.tanB=1

adding "1" on both sides

1+ tanA+tanB+tanA.tanB=1+1

(1 + tanA)+tanB(1+tanA).=2

(1+tanA)(1+tanB)=2 Hence proved .

Answer 2

Given:-

• A + B = 45°

Apply tan on both sides.

⟶ tan (A + B) = tan 45°

⟶ tan (A + B) = (tan A + tan B) / 1 - tan A tan B

⟶ tan 45° = 1

So,

⟶ tan A + tan B / 1 - tan A tan B = 1

⟶ tan A + tan B = 1 - tan A tan B

⟶ tan A + tan B + tan A tan B = 1

Adding 1 on both sides we get,

⟶ tan A + tan A tan B + tan B + 1 = 1 + 1

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

⟶ (1 + tan B)(1 + tan A) = 2

⟶ (1 + tan A)(1 + tan B) = 2

Hence, Proved !

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Additional Information:-

• sin (A + B) = sin A cos B + cos A sin B

• cos (A + B) = cos A cos B - sin A sin B.

• sin (A - B) = sin A cos B - cos A sin B

• cos (A - B) = cos A cos B + sin A sin B.

• sin 2A = 2 sin A cos A

• cos 2A = cos² A - sin² A.