Question

prove (1+TanA)(1+tanB)=2,A+B=45°​

Answer 1

Given, A + B = 45°

⇒ tan(A + B) = tan45°

⇒ (tanA + tanB)/(1 - tanAtanB) = 1

⇒ tanA + tanB = 1 - tanAtanB

⇒ tanA + tanB + tanAtanB = 1

∴ (1 + tanA)(1 + tanB)

⇒ 1 + tanB + tanA + tanAtanB

⇒ 1 + 1

⇒ 2            proved

Answer 2

Answer:-

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.