If tan A - a - a -1 and tan B -1-2 a -1- then tha value of A - B isA- 0 B- -2C- -3D- -4

The correct option is D π/4tan(A+B)= tanA+tanB/1 tanA tanB = a/a+1+ 1/2a+1/1 a/(a+1)(2a+1) = 2a^2 +a+a+1/2a^2+3a+1 a = 2a^2+2a+1/2a^2+2a+1= 1 Therefore, A+B= t ...

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

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tan A = a ...

Question

If $t a n A = \frac{a}{a + 1}$ and $t a n B = \frac{1}{2 a + 1},$ then tha value of A+B is

\frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $\frac{π}{4}$
$t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$
$= \frac{\frac{a}{a + 1} + \frac{1}{2 a + 1}}{1 - \frac{a}{(a + 1) (2 a + 1)}}$
$= \frac{2 a^{2} + a + a + 1}{2 a^{2} + 3 a + 1 - a}$
$= \frac{2 a^{2} + 2 a + 1}{2 a^{2} + 2 a + 1} = 1$
Therefore, $A + B = t a n^{- 1} (1) = \frac{π}{4} .$

Suggest Corrections

Similar questions

(i) If $t a n A = \frac{5}{6}$ and $t a n B = \frac{1}{11},$ prove that A+B= $\frac{π}{4}$
(ii) If $t a n A = \frac{m}{m - 1}$ and $t a n B = \frac{m}{2 m - 1}$
then prove that $A - B = \frac{π}{4}$