Question

tanx = n/ n+1 ; tan y=1 / 2n+1 find tan (x+y) ​

Answer 1

Answers :

Answer is in attached file.

Answer 2

Step-by-step explanation:

$\large{ \green{ \rm \: GIVEN : - }}$

$\sf \: \tan(x) = \frac{n}{n + 1}$

$\sf \: \tan( y) = \frac{1}{2n + 1}$

$\large{ \green{ \rm \: TO \: \: FIND : - }}$

$\sf \: \tan(x + y)$

$\large{ \green{ \rm \: SOLUTION : - }}$

$\sf \: Using \: \: the \: \: formula$

$\sf \: \tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x). \tan(y) }$

$\sf \: Now \: \: putting \: \: the \: \: value : -$

$\implies\large \sf \: \tan(x + y) = \frac{ \frac{n}{n + 1} + \frac{1}{2n + 1} }{1 - ( \frac{n}{n + 1})( \frac{1}{2n + 1} )}$

$\implies\large \sf \: \tan(x + y) = \frac{ \frac{n(2n + 1) + n + 1}{(n + 1)(2n + 1)} }{1 - ( \frac{n}{(n + 1)(2n + 1)})}$

$\implies \large \sf \: \tan(x + y) = \frac{ \frac{2n {}^{2} + n + n + 1}{(n + 1)(2n + 1)} }{ ( \frac{(n + 1)(2n + 1) - n}{(n + 1)(2n + 1)})}$

$\implies \large \sf \: \tan(x + y) = \frac{ \frac{2n {}^{2} + 2 n + 1}{(n + 1)(2n + 1)} }{ ( \frac{2 {n}^{2} + n + 2n + 1 - n}{(n + 1)(2n + 1)})}$

$\implies \sf \: \tan(x + y) = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \frac{ (2n^{2} + 2 n + 1)}{\cancel{(n + 1)(2n + 1)}} \times \frac{\cancel{(n + 1)(2n + 1)}}{ (2 {n}^{2} + \cancel{ n} + 2n + 1 -\cancel{ n} )}$

$\implies \sf \tan(x + y) = \frac{2 {n}^{2} + 2n + 1 }{2 {n}^{2} + 2n + 1}$

$\implies \sf \: \tan(x + y) = 1$