Question

if A+B =45° then prove that (1+tanA) (1+tanB)=2 and hence deduce that tan22 1/2°​

Answer 1

Step-by-step explanation:

Given:-

A+B =45°

To find:-

Prove that (1+tanA) (1+tanB)=2 and hence deduce that tan22 1/2°

Solution:-

Given that : A+B = 45°

On taking Tan both sides then

=> Tan (A+B)=Tan 45°

=>( Tan A+ Tan B)/(1-Tan A TanB) = 1

=> Tan A+ Tan B = 1- Tan A Tan B

=> Tan A+ Tan B + Tan A Tan B = 1 ---(1)

LHS:

(1+ Tan A)(1+TanB)

=>1(1+ Tan B) + Tan A(1+TanB)

=> 1+ Tan B + Tan A+ Tan A Tan B

=> 1+ ( Tan A Tan B + Tan A Tan B)

=> 1+1 ( from (1))

=> 2

=> RHS

LHS= RHS

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

Hence, Proved

We know that

Tan A/2 = √[(1- Cos A)/(1+CosA)]

Take A = 45° then A/2 = 22 1/2°

=> Tan 45°/2

=>√[(1- Cos 45°)/(1+Cos45°)]

=>√[(1-(1/√2))/(1+(1/√2))]

=>√[{(√2-1)/√2}/{(√2+1)/√2}]

=>√[(√2-1)/(√2+1)

On Rationalising the denominator

=>√[(√2-1)(√2-1)/(√2+1)(√2-1)]

=>√[(√2-1)^2/(2-1)]

=>√(√2-1)^2/1

=>√(√2-1)^2

=>√2-1

Tan 22 1/2° = √2-1

Answer:-

Tan 22 1/2° = √2-1

Used formulae:-

• Tan (A+B)=( Tan A+ Tan B)/(1-Tan A TanB)

• Tan A/2 = √[(1- Cos A)/(1+CosA)]

• Tan 45°=1

• Cos 45° = 1/√2