Question

$\frac{(1 + \tan \: 13)(1 + \tan \: 32) }{(1 + \tan \: 12)(1 + \tan \: 33) } =$


A) 1
B) 2
C) 3
D) 4
​

Answer 1

To Find :-

• (1 + Tan13°)(1+Tan32°) / (1+Tan12°)(1+Tan33°)

Solution :-

we know That, if A + B = 45° ,

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

→ in Numerator we have A = 13° , B = 32°

and, A + B = 13+32 = 45° .

→ Similarly, in Denominator, A = 12° , B = 33°

and, again A + B = 12+33 = 45° .

So, We can say That,

→ (1 + Tan13°)(1+Tan32°) = 2

→ (1+Tan12°)(1+Tan33°) = 2

Putting Both Values we get,

→ 2/2

→ 1 (Ans).

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Extra Knowledge :-

Lets Try to Prove This Formula now :-

→ A + B = 45°

→ tan(A + B) = tan45°

→ (tanA + tanB)/(1 - tanA*tanB) = 1

→ (tanA + tanB) = (1 - tanA*tanB)

→ tanA + tanB + tanA*tanB = 1

Adding 1 both sides ,

→ (tanA + 1) + (tanB + tanA*tanB) = 2

→ (tanA + 1) + tanB(1 + tanA) = 2

→ (tanA + 1)( 1 + tanB) = 2

→ (1 + tanA)(1 + tanB) = 2 (Proved).

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Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Problem-}}}}}}$

$\frac{(1 + \tan \: 13)(1 + \tan \: 32) }{(1 + \tan \: 12)(1 + \tan \: 33) } =$

____________________________________________

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer:-}}}}}}$

We know that if,

(A+B)= 45°

Than,

(1+tanA + tanB) =2

___________________________________________

In the position of numerator we have,

A=13°

B =32°

=> A+B = 13°+32°=45°

___________________________________________

In the position of denominator we have,

A=12° B =33°

=> A + B =45°

____________________________________________

We can also say that,

=> (1+tan12°)+(1+tan33°)=2

____________________________________________

putting both values together,

=> 2/2

=>1

Hence,the required answer is 1