Question

tanA=1/2and tanB=1/3then tan(2A+B)? ​

Answer 1

Answer:

tanA=1/2and tanB=1/3

tan(2A+B)=tan2A+tanB

• =tan2A +1/3
• =2(1/2)+1/3
• =1+1/3
• =3+1/3
• =4/3

Answer 2

Answer:

$\large\boxed{\sf{3}}$

Step-by-step explanation:

Let, $\angle A = \alpha$ and$\ angle B = \beta$

Therefore, we have,

$\tan( \alpha ) = \frac{1}{2}$

$= > \tan(2 \alpha ) = \frac{2 \tan( \alpha ) }{1 - { \tan }^{2} \alpha } \\ \\ = > \tan(2 \alpha ) = \frac{2 \times \frac{1}{2} }{1 - {( \frac{1}{2} )}^{2} } \\ \\ = > \tan(2 \alpha ) = \frac{1}{1 - \frac{1}{4} } \\ \\ = > \tan(2 \alpha ) = \dfrac{1}{ \frac{3}{4} } \\ \\ = > \sf{ \tan(2 \alpha ) = \frac{4}{3} }$

And, we have from question,

$\sf{\tan( \beta ) = \dfrac{1}{3} }$

Now, to find the value of,

$\tan(2 \alpha + \beta ) = \frac{ \tan(2 \alpha ) + \tan( \beta ) }{1 - \tan(2 \alpha ) \tan( \beta ) } \\ \\ = > \tan(2 \alpha + \beta ) = \frac{ \frac{4}{3} + \frac{1}{3} }{1 - (\frac{4}{ 3} \times \frac{1}{3}) } \\ \\ = > \tan(2 \alpha + \beta ) = \frac{ \frac{5}{3} }{1 - \frac{4}{9} } \\ \\ = > \tan(2 \alpha + \beta ) = \frac{ \frac{5}{3} }{ \frac{5}{9} } \\ \\ = > \tan(2 \alpha + \beta ) = \frac{9}{3} \\ \\ = > \sf{\tan(2 \alpha + \beta ) = 3}$

Hence, tan(2A+B) = 3