Question

if tan(alpha+theta) = ntan (alpha-theta) then show that : (n+1)sin 2theta=(n-1)sin 2alpha

Answer 1

Given: $\tan(\alpha + \Theta) = n \tan(\alpha - \theta)$

To Prove: $(n+1) \sin2\theta = (n-1)\sin2\alpha$

proof:

Consider $\tan(\alpha + \Theta) = n \tan(\alpha - \theta)$

By using the trigonometric identities

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

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

Therefore,

$\frac{\tan \alpha +\tan \Theta }{1-\tan\alpha \tan \Theta }= n[ \frac{\tan \alpha -\tan \Theta }{1+\tan\alpha \tan \Theta }]$

$({\tan \alpha +\tan \Theta })({1+\tan\alpha \tan \Theta })= ({1-\tan \alpha \tan \Theta })({n \tan\alpha -n \tan \Theta })$

$\tan \alpha +\tan ^{2}\alpha \tan \Theta +\tan \Theta +\tan\alpha\tan^{2} \Theta = n\tan \alpha-n \tan \Theta - n \tan^{2}\alpha \tan \Theta +n \tan \alpha \tan ^{2}\Theta$

$\tan \theta(1+\tan ^{2}\alpha) +n\tan \Theta(1+\tan^{2}\alpha) = n\tan \alpha(1+\tan ^{2}\Theta )- \tan \alpha (1+\tan^{2}\theta)$

$(n+1)\tan \theta(1+\tan ^{2}\alpha) = (n-1)\tan \alpha(1+\tan ^{2}\Theta )$

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

Multiplying both sides of the above equation by 2,

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

By applying the trigonometric identities, we get

$(n+1)\sin 2\theta = (n-1)\sin 2\alpha$

Hence, proved.