If tan A = - 1 2 and tan B = - 1 3 , then A + B = [IIT 1967; MNR 1987; MP PET 1989] Options: A . &nbsp π 4 B . &nbsp 3 π 4 C . &nbsp 5 π 4 D . &nbsp None 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 1

Discussion Forum : Trigonometric Functions

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 .

\frac{π}{4}

B .

\frac{3 π}{4}

C .

\frac{5 π}{4}

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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Discussion Forum : Trigonometric Functions

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