Question

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Question for maths expert......

Answer 1

tan(alpha+beta)= tan alpha + tan beta/1- tan alpha tan beta

= m/m+1 + 1/2m +1 / 1 - m/(m+1)(2m+1)

= 2m^2 + m + m+ 1 / (m+1)( 2m+1) -m

= 2m^2 + 2m + 1)/ 2m^2 + m + 2m +1 -m

= 2m^2 + 2m +1)/ 2m^2 + 2m +1

= 1

So alpha + beta = pie/4


✌✌✌Dr.Dhruv✌✌✌✌✌✌

✌✌✌✌Senior chairman of Resdnick✌✌✌✌

Answer 2

Hi there !
Here's the answer :

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°

Taking alpha as a an Beta as b

Given,
$tan\: a= \frac{m}{m+1}$

$tan\: b= \frac{1}{2m+1}$


We have,
$tan(a+b) = \frac{tan\: a+tan\: b}{1-tan\: a.tan\: b}$

=$\frac{\frac{m}{m+1}+\frac{1}{2m+1}}{1-(\frac{m}{m+1})(\frac{1}{2m+1})}$

=$\frac{m(2m+1)+(m+1)}{(m+1)(2m+1)-m}$

= $\frac{2m^{2}+2m+1}{2m^{2}+3m+1-m}$


= $\frac{2m^{2}+2m+1}{2m^{2}+2m+1}$

= $1$



$tan (a+b) = 1$

=> $(a+b) = tan^{-1}(1)$

=> $(a+b) = tan^{-1}(tan\: \frac{\pi}{4})$


•°•
$(a+b) = \frac{\pi}{4}$