If tanA - 1-2-tanB - 1-3- then tan A - B -

[ tanA = 12,tanB = 13; tan( A + B) = tanA + tanB1 tanAtanB; = 1/2 + 13/1 12.13 = 1 ] ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Domain

If tanA = 1...

Question

If $t a n A = \frac{1}{2}, tan B = \frac{1}{3},$ then $t a n (A + B) =$

Open in App

Solution

$\begin{matrix} t a n A = \frac{1}{2}, tan B = \frac{1}{3} t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} . \frac{1}{3}} = 1 \end{matrix}$

Suggest Corrections

Similar questions

Q. If

tan A = - \frac{1}{2}

and

tan B = - \frac{1}{3}

then

A + B =

Q. If

tan A = \frac{1}{35} . tan B = \frac{18}{17}

, then

tan (A - B)

If A and B are acute angles such that tan $A = \frac{1}{2}, t a n B = \frac{1}{3}$ and tan $(A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ ,find A + B

Q. If

tan A = - \frac{1}{2}

and

tan B = - \frac{1}{3}

. (where

A, B > 0

), then

A + B

can be

Q. If

tan A = \frac{1}{\sqrt{3}}, tan B = \sqrt{3}

then find the value of A

+

Join BYJU'S Learning Program

If tanA=12,tanB=13, then tan(A+B)=

tanA=12,tanB=13tan(A+B)=tanA+tanB1−tanAtanB=12+131−12.13=1

If $t a n A = \frac{1}{2}, tan B = \frac{1}{3},$ then $t a n (A + B) =$

$\begin{matrix} t a n A = \frac{1}{2}, tan B = \frac{1}{3} t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} . \frac{1}{3}} = 1 \end{matrix}$