# TRIGONOMETRIC RATIOS OF COMPOUND ANGLES (PART-1)

$\,1.\,$ Prove that, $\,\,\tan(A+B)-\tan(A-B)=\frac{\sin 2B}{\cos^2B-\sin^2A}.$

Sol.  We have, $\,\,\tan(A+B)-\tan(A-B)\\=\frac{\sin(A+B)}{\cos(A+B)}-\frac{\sin(A-B)}{\cos(A-B)}\\=\frac{\sin(A+B)\cos(A-B)-\cos(A+B)\sin(A-B)}{\cos(A+B)\cos(A-B)}\\=\frac{\sin(A+B-A+B)}{\cos^2B-\sin^2A}\\=\frac{\sin2B}{\cos^2B-\sin^2A}\,\,\,\text{(proved)}$

$\,2.\,$ If  $\,\,\,\tan\beta=\frac{n\sin\alpha \cos\alpha }{1-n\sin^2\alpha },\,\,$ show that , $\,\,\tan(\alpha -\beta)=(1-n)\tan\alpha .$

Sol. $\,\,\tan(\alpha -\beta)\\=\frac{\tan\alpha -\tan\beta}{1+\tan\alpha \tan\beta}\\=\frac{\frac{\sin\alpha }{\cos\alpha }-\frac{n\sin\alpha \cos\alpha }{1-n\sin^2\alpha }}{1+\frac{\sin\alpha }{\cos\alpha }. \frac{n\sin\alpha \cos\alpha }{1-n\sin^2\alpha }}\\=\frac{\sin\alpha(1-n\sin^2\alpha)-\cos\alpha \times n\sin\alpha\cos\alpha}{\cos\alpha(1-n\sin^2\alpha)+n\sin^2\alpha\cos\alpha}\\=\frac{\sin\alpha-n\sin^3\alpha-n\sin\alpha\cos^2\alpha}{\cos\alpha-n\sin^2\alpha\cos\alpha+n\sin^2\alpha\cos\alpha}\\=\frac{\sin\alpha-n\sin\alpha(\sin^2\alpha+\cos^2\alpha)}{\cos\alpha}\\=\frac{\sin\alpha-n\sin\alpha\times 1}{\cos\alpha}\\=\frac{(1-n)\sin\alpha}{\cos\alpha}\\=(1-n)\tan\alpha\quad\text{(proved)}$

$\,3.\,$ If $\,\,m\tan(\theta-30 ^{\circ})=n\tan(\theta+120 ^{\circ}),\,\,$ show that , $\,\,2\cos2\theta=\frac{m+n}{m-n}.$

Sol. $\,\,m\tan(\theta-30 ^{\circ})=n\tan(\theta+120 ^{\circ}) \\ \Rightarrow \frac mn=\frac{\tan(\theta+120 ^{\circ})}{\tan(\theta-30 ^{\circ})} \\ \Rightarrow \frac mn=\frac{\sin(\theta+120 ^{\circ})\cos(\theta-30 ^{\circ})}{\cos(\theta+120 ^{\circ})\sin(\theta-30 ^{\circ})}\\ \Rightarrow \frac{m+n}{m-n}\\=\frac{\sin(\theta+120 ^{\circ})\cos(\theta-30 ^{\circ})+\cos(\theta+120 ^{\circ})\sin(\theta-30 ^{\circ})}{\sin(\theta+120 ^{\circ})\cos(\theta-30 ^{\circ})-\cos(\theta+120 ^{\circ})\sin(\theta-30 ^{\circ})}\\=\frac{\sin(\theta+120^{\circ}+\theta-30^{\circ})}{\sin(\theta+120^{\circ}-\theta+30^{\circ})}\\=\frac{\sin(90^{\circ}+2\theta)}{\sin150^{\circ}}\\=\frac{\cos2\theta}{\sin(90^{\circ}+60^{\circ})}\\=\frac{\cos2\theta}{\cos60^{\circ}}\\=\frac{\cos2\theta}{\frac 12}\\=2\cos2\theta \\ \therefore 2\cos2\theta=\frac{m+n}{m-n}\quad\text{(showed)}$

$\,4.\,$ If $\,\,\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac 32,\,\,$ show that , $\,\,\cos\alpha+\cos\beta+\cos\gamma=0\,\,$ and $\,\,\sin\alpha+\sin\beta+\sin\gamma=0.$

Hence, deduce that , $\,\,\cos(\beta-\gamma)=\cos(\gamma-\alpha)=\cos(\alpha-\beta)=-\frac 12 .$

Sol. We have, $\,\,\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac 32\\ \Rightarrow 2\cos\beta\cos\gamma+2\sin\beta\sin\gamma+2\cos\gamma\cos\alpha\\+2\sin\gamma\sin\alpha+2\cos\alpha\cos\beta+2\sin\alpha\sin\beta \\+3=0\\ \Rightarrow 2\cos\beta\cos\gamma+2\sin\beta\sin\gamma+2\cos\gamma\cos\alpha\\+2\sin\gamma\sin\alpha+2\cos\alpha\cos\beta+2\sin\alpha\sin\beta\\+(\sin^2\alpha+\cos^2\alpha)+(\sin^2\beta+\cos^2\beta)\\+(\sin^2\gamma+\cos^2\gamma)=0 \\ \Rightarrow (\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\beta\cos\gamma\\+2\cos\gamma\cos\alpha+2\cos\alpha\cos\beta)\\+(\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\\+2\sin\beta\sin\gamma+2\sin\gamma\sin\alpha)=0 \\ \Rightarrow (\cos\alpha+\cos\beta+\cos\gamma)^2\\+(\sin\alpha+\sin\beta+\sin\gamma)^2=0\rightarrow(1)$

So, $\,\,(1)\,\,$ is possible only when  $\,\,\cos\alpha+\cos\beta+\cos\gamma=0\rightarrow(2) \\ \sin\alpha+\sin\beta+\sin\gamma=0\rightarrow(3)$

From $\,\,(2)\,$ we get, $\,\,\cos\alpha+\cos\beta=-\cos\gamma \\ \Rightarrow (\cos\alpha+\cos\beta )^2=\cos^2\gamma \rightarrow(4)$

From $\,\,(3)\,$ we get, $\,\,\sin\alpha+\sin\beta=-\sin\gamma \\ \Rightarrow (\sin\alpha+\sin\beta )^2=\sin^2\gamma \rightarrow(5)$

Adding $\,(4)\,$ and $\,(5),\,$ we get,

$\,\,(\cos\alpha+\cos\beta )^2+(\sin\alpha+\sin\beta )^2\\=\cos^2\gamma+\sin^2\gamma \\ \Rightarrow \cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta+\sin^2\alpha\\+\sin^2\beta+2\sin\alpha\sin\beta=\cos^2\gamma+\sin^2\gamma \\ \Rightarrow (\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)\\+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)=1 \\ \Rightarrow 1+1+2\cos(\alpha-\beta)=1 \\ \Rightarrow \cos(\alpha-\beta)=-\frac 12.$

Similarly, we can prove that $\,\,\cos(\gamma-\alpha)=\cos(\beta-\gamma)=-\frac 12 .$

$\,5.\,$ If $\,\,\alpha \neq \beta\,\,$ and $\,\,a\tan\alpha+b\tan\beta=(a+b)\tan\frac{\alpha+\beta}{2},\,\,$ show that, $\,\,\frac{\cos\alpha}{\cos\beta}=\frac ab.$

Sol. We have,  $\,\,a\tan\alpha+b\tan\beta=(a+b)\tan\frac{\alpha+\beta}{2}\\ \Rightarrow a\left(\tan\alpha-\tan\frac{\alpha+\beta}{2}\right)=b\left(\tan\frac{\alpha+\beta}{2}-\tan\beta\right) \\ \Rightarrow a\left(\frac{\sin\alpha}{\cos\alpha}-\frac{\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha+\beta}{2}}\right)=b\left(\frac{\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha+\beta}{2}}-\frac{\sin\beta}{\cos\beta}\right)\\ \Rightarrow a\times \frac{\sin\left(\alpha-\frac{\alpha+\beta}{2}\right)}{\cos\alpha\cos\frac{\alpha+\beta}{2}}=b\times \frac{\sin\left(\frac{\alpha+\beta}{2}-\beta\right)}{\cos\beta\cos\frac{\alpha+\beta}{2}} \\ \Rightarrow \frac{a\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos\alpha}=\frac{b\sin\left(\frac{\alpha-\beta}{2}\right)}{\cos\beta} \\ \Rightarrow \frac{\cos\alpha}{\cos\beta}=\frac ab\quad\text{(showed)}$