Question

"Question40
If A and B are acute angles such that tan A = 1/3 , tan B =1/2 and tan (A+B ) = tanA+tanB / 1 - tan A tan B, show that A + B = 45°
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289"

Answer 1

Hey there!

tanA = 1/3
tanB = 1/2 .

First, Let's see tan45° = 1 [ Trigonometry of particular angles ]

Now, We have to prove that tan(A+B) = tan(45) = 1 .

So, Now

tan ( A + B)

=> tanA + tanB / 1 - tanA * tanB

=> 1/3 + 1/2 / 1 - (1/2)(1/3)

=> 2 + 3 / 6 / 1 - 1/6

=>[ 5/6 ] / 5/6

=> 1 .

So, tan(A + B) = 1

We know that, 1 = tan45 , Replace 1 with tan45

tan(A + B) = tan( 45)

Now, Comparing the equations.

A + B = 45°

Hope helped!

Answer 2

Hey mate !!

Here's the answer !!

Given:

Tan A = 1 / 3
Tan B = 1 / 2

Tan ( A + B ) = Tan A + Tan B / 1 - Tan A . Tan B

To show :

A + B = 45°

Proof :

We know the value of A and B. So let us substitute them in the given equation to find what Tan ( A + B )  equals to.

Substituting them we get,

Tan ( A + B ) = Tan 1 / 2 + Tan 1 / 3 / 1 - Tan 1 / 2 . Tan 1 / 3

Considering the RHS we get,

LHS = Tan ( 1 / 2 + 1 / 3 ) / 1 -  Tan ( 1 / 2 * 1 / 3 )

LHS = Tan ( 3 + 2 / 3 * 2 ) / 1 - Tan 1 / 6

LHS = Tan ( 5 / 6 ) / Tan ( 5 / 6 )

=> LHS = 1.

=> Tan ( A + B ) = 1

We know that Tan 45° = 1.

Hence substitute 1 as 45 in the above equation. We get,

Tan ( A + B ) = Tan 45

Tan gets cancelled on both sides. Hence,

A + B = 45°.

Hence proved !!

Hope my answer helped you mate !!

Cheers !!