help me to show that $\frac{\tan(\alpha-\beta)+\tan\beta}{1-\tan(\alpha-\beta)\tan\beta}=\frac{m^2-n^2}{2mn}$

Question

help me to solve this problem.How can I approach? If $\sin\alpha=\frac{m^2-n^2}{m^2+n^2}$ then show that

$$\frac{\tan(\alpha-\beta)+\tan\beta}{1-\tan(\alpha-\beta)\tan\beta}=\frac{m^2-n^2}{2mn}$$

Answer 1

What you should know to solve this problem is that,

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$

In your case you get,

$$\frac{\tan(\alpha-\beta)+\tan\beta}{1-\tan(\alpha-\beta)\tan\beta}=\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$

Now you know what $\sin\alpha$ is, but can you also try to find what is $\cos \alpha$ ?

Answer 2

since sin X= perpendicular/hypotenuse

let ABC be a right angled triangle right-angled at B.

then perpendicular side= AB= m^2-n^2

and hypotenuse= AC= m^2+n^2

and AC^2= AB^2 + BC^2

so, after solving, BC=base= 2mn

now, tan X= perpendicular/base

and tan(A+B)= (tan A+ tan B)/ (1- tanA tanB)

here, A= α−β, B= β

thus,(tan(α−β)+tanβ)/(1−tan(α−β)tanβ)=tanα

hence tan α= perpendicular/base = (m^2-n^2)/2mn