Question

If tan a = m/(m+1)
tan b = 1/(2m+1)

prove that tan(a+b) does not depend on the value of m .

Answer 1

Answer:

$tana=\dfrac{m}{m+1}\\\\tanb=\dfrac{1}{2m+1}\\\\tan(a+b)=\dfrac{tana+tanb}{1-tanatanb}\\\\\implies tan(a+b)=\dfrac{\dfrac{m}{m+1}+\dfrac{1}{2m+1}}{1-\dfrac{m}{m+1}\times\dfrac{1}{2m+1}}\\\\\\\implies tan(a+b)=\dfrac{\dfrac{m(2m+1)+m+1}{(m+1)(2m+1)}}{\dfrac{(m+1)(2m+1)-m}{(m+1)(2m+1)}}\\\\\\\implies tan(a+b)=\dfrac{2m^2+m+m+1}{2m^2+m+2m+1-m}\\\\\implies tan(a+b)=\dfrac{2m^2+2m+1}{2m^2+2m+1}\\\\\implies tan(a+b)=1$

Explanation:

Use the formulas of trigonometry :

$tan(X+Y)=\dfrac{tanX+tanY}{1-tanXtanY}$

Calculate the value of tan ( a + b ) using the given values of tan a and tan b .

We see that tan ( a + b ) yields 1 .

Hence the solution of tan ( a + b ) does not depend on the value of m.

In simple words , for any value of m , tan ( a + b ) will be 1 .

Answer 2

hope it help..........