Question

if (sin theta + alpha)/ (cos theta - alpha) = (1- m)/(1+m). Prove that tan (pi/4 - theta) tan (pi/4 - alpha) = m

Answer 1

Given: $\frac{sin (\theta + \alpha)} {cos (\theta - \alpha)} = \frac{1- m}{1+m}$

To prove: $\tan (\frac{\pi}{4} - \theta) \tan (\frac{\pi}{4} - \alpha) = m$

Consider $\tan (\frac{\pi}{4} - \theta) \tan (\frac{\pi}{4} - \alpha)$

Consider the trigonometric identity

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

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

Consider $\tan (\frac{\pi}{4} - \theta) \tan (\frac{\pi}{4} - \alpha)$

= $[\frac{\tan \frac{\Pi }{4} - \tan \theta}{1 + \tan \frac{\Pi }{4} \tan \theta}] [\frac{\tan \frac{\Pi }{4} - \tan \alpha}{1 + \tan \frac{\Pi }{4} \tan \alpha}]$

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

Converting tan in the form of sin and cos, we get

=$(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}) (\frac{\cos \alpha-\sin \alpha}{\cos \alpha+\sin \alpha})$

= $(\frac{\cos \theta\cos \alpha -\cos \theta \sin \alpha - \sin \theta \cos  \alpha+\sin \theta \sin  \alpha}{\cos \theta\cos \alpha +\cos \theta \sin \alpha + \sin \theta \cos  \alpha+\sin \theta \sin  \alpha})$

=$(\frac{\cos (\theta -\alpha) -\sin( \theta +\alpha)}{\cos (\theta -\alpha) +\sin( \theta +\alpha)})$

Dividing numerator and denominator by $\cos (\theta - \alpha)$,

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

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

= $\frac{1+m-1+m}{1+m+1-m}$

= $\frac{2m}{2}$

= m

Hence, proved.

Answer 2

Answer:

Given Equation is

$\frac{sin\,(\theta+\alpha)}{cos\,(\theta-\alpha)}=\frac{1-m}{1+m}$

To Prove: $tan\,(\frac{\pi}{4}-\theta)\:tan\,(\frac{\pi}{4}-\alpha)=m$

We use the following trigonometric identity  ,

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

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

Consider

$tan(\frac{\pi}{4}-\theta)\:tan(\frac{\pi}{4}-\alpha)$

$=(\frac{tan\,\frac{\pi}{4}-tan\,\theta}{1+tan\,\frac{\pi}{4}\:tan\,\theta}) (\frac{tan\,\frac{\pi}{4}-tan\,\alpha}{1+tan\,\frac{\pi}{4}\:tan\,\alpha})$

$=(\frac{1-tan\,\theta}{1+tan\,\theta})(\frac{1-tan\,\alpha}{1+tan\,\alpha})$

$=(\frac{cos\,\theta-sin\,\theta}{cos\,\theta+sin\,\theta})(\frac{cos\,\alpha-sin\,\alpha}{cos\,\alpha+sin\,\alpha})$

$=(\frac{cos\,\theta\:\,cos\,\alpha-cos\,\theta\:\,sin\,\alpha-sin\,\theta\:\,cos\,\alpha+sin\,\theta\:\,sin\alpha}{cos\theta\:\,cos\alpha+cos\,\theta\:\,sin\alpha+sin\,\theta\:\,cos\,\alpha+sin \,\theta\:\,sin\,\alpha})$

$=(\frac{cos(\theta-\alpha)-sin(\theta+\alpha)}{cos(\theta-\alpha)+sin(\theta +\alpha)})$

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

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

$=\frac{1+m-1+m}{1+m+1-m}$

$=\frac{2m}{2}$

$=m$

Hence, proved.