Question

if tan A =under root 3 and tan B=1/under root 3 then find the value of cot (A+B)

Answer 1

Given, tan A=root3 and tan b=1/root 3

tan A=60degree and tanB=30 degree

So, A=60 degree and B = 30 degree.

cot (A+B)=cot(60+30)=city 90degree .

cot90degree =0

Answer 2

Given:

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

To find: the value of $\[\cot \left( {A + B} \right)\]$.

Solution:

Find the value of  $\[\cot A\]$.

$\[\begin{array}{l}\tan A = \sqrt 3 \\ \Rightarrow \cot A = \frac{1}{{\tan A}}\end{array}\]$

$\[ \Rightarrow \cot A = \frac{1}{{\sqrt 3 }}\]$

Find the value of $\[\cot B\]$.

$\[\begin{array}{l}\tan B = \frac{1}{{\sqrt 3 }}\\ \Rightarrow \cot B = \frac{1}{{\tan B}}\end{array}\]$

$\[ \Rightarrow \cot B = \frac{1}{{\frac{1}{{\sqrt 3 }}}}\]$

$\[ \Rightarrow \cot B = \sqrt 3 \]$

Find the value of $\[\cot \left( {A + B} \right)\]$.

Apply, $\[\cot \left( {A + B} \right) = \frac{{\cot A\cot B - 1}}{{\cot B + \cot A}}\]$.

Substitute the values of cot A and cot B in above formulae.

$\[ \Rightarrow \cot \left( {A + B} \right) = \frac{{\sqrt 3 \times \frac{1}{{\sqrt 3 }} - 1}}{{\sqrt 3 + \frac{1}{{\sqrt 3 }}}}\]$

$\[ \Rightarrow \cot \left( {A + B} \right) = \frac{0}{{\sqrt 3 + \frac{1}{{\sqrt 3 }}}}\]$

$\[ \Rightarrow \cot \left( {A + B} \right) = 0\]$

Hence, the value of $\[\cot \left( {A + B} \right)\]$ is 0.