If tan A = - 1 2 and tan B = - 1 3 , then A + B = [IIT 1967; MNR 1987; MP PET 1989] Options: A . &nbspπ4 B . &nbsp3π4 C . &nbsp5π4 D . &nbspNone of these

Answer: Option B : B We have tan A = - 1 2 and tan B = - 1 3 Now, tan(A + B) = t a n A + t a n B 1 − t a n A t a n B = − 1 2 − 1 3 1 − 1 2 . 1 3 = -1 ⇒ tan(A + B) = tan 3 π 4 . Hence, A + B = 3 π 4 .

If tan A = - 12 and tan B = - 13, then A + B = [IIT 1967; MNR 198

Question

If tan A = -

\frac{1}{2}

and tan B = -

\frac{1}{3}

, then A + B =
[IIT 1967; MNR 1987; MP PET 1989]

Options:

A . π4

B . 3π4

C . 5π4

D . None of these

Answer: Option B
:
B
We have tan A = -

\frac{1}{2}

and tan B = -

\frac{1}{3}

Now, tan(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

\Rightarrow

tan(A + B) = tan

\frac{3 π}{4}

. Hence, A + B =

\frac{3 π}{4}

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