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Question Number 196886 by Amidip last updated on 02/Sep/23

Answered by MM42 last updated on 02/Sep/23

$$tan(\alpha +\beta )=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }=\frac{p}{q-1}$$$$\Rightarrow \frac{p}{q-1}=\frac{{sin}^2(\alpha +\beta )}{sin(\alpha +\beta )cos(\alpha +\beta )}$$$$\Rightarrow psin(\alpha +\beta )cos(\alpha +\beta )=(q-1){sin}^2(\alpha +\beta ))$$$$\Rightarrow {sin}^2(\alpha +\beta )+psin(\alpha +\beta )cos(\alpha +\beta )={qsin}^2(\alpha +\beta )✓$$

Answered by HeferH last updated on 02/Sep/23

$$(x-\tan a)(x-\tan b)=0$$$$x^2-x(\tan a+\tan b)+\tan a\tan b=0$$$$p=-(\tan a+\tan b)$$$$q=\tan a\tan b$$$${\sin }^2(a+b)+p\sin (a+b)\cos (a+b)+q\cos {}^2(a+b)=q$$$${\sin }^2(a+b)+p\sin (a+b)\cos (a+b)=q\sin {}^2(a+b)$$$$1+p\cot (a+b)=q$$$$\cot (a+b)=\frac{q-1}{p}$$$$\frac{q-1}{p}=\frac{1-\tan a\tan b}{(\tan a+\tan b)}=\tan {}^{-1}(a+b)=\cot (a+b)$$

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