Differentiation (Part-38) | S N De

In the previous article , we have discussed Short Ans Type Questions (from Question No.6 to 8)of the Chapter : TRANSFORMATIONS OF SUMS AND PRODUCTS (Part-5) as a part of our S.N.Dey Math Solution series. In this article, we will continue our journey to solve few more mathematical problems.  

$9(i)\,\,$ If $\,\,13\theta=\pi,\,\,$ show that, $\,\,\cos3\theta+\cos5\theta+2\cos\theta\cos9\theta=0.$

Sol. $\,\cos3\theta+\cos5\theta+2\cos\theta\cos9\theta\\=2\cos\frac{5\theta+3\theta}{2}\cos\frac{5\theta-3\theta}{2}+2\cos\theta\cos9\theta\\=2\cos4\theta\cos\theta+2\cos\theta\cos9\theta\\=2\cos\theta(\cos4\theta+\cos9\theta)\\=2\cos\theta \times 2\cos\frac{9\theta+4\theta}{2}\cos\frac{9\theta-4\theta}{2}\\=2\cos\theta \times 2\cos\frac{13\theta}{2}\cos\frac{5\theta}{2}\\=4\cos\theta \cos\frac{5\theta}{2}\cos\frac{\pi}{2}\,\,[\text{Since,}\,\,\,13\theta=\pi]\\=0\,\,[\,\text{Since,}\,\,\cos\frac{\pi}{2}=0]$

$9(ii)\,\,$ If $\,\,\sin2\alpha=4\sin2\beta,\,$ show that, $5\tan(\alpha-\beta)=3\tan(\alpha+\beta)$

Sol. $\,\,\sin2\alpha=4\sin2\beta\\ \Rightarrow \frac{\sin2\alpha}{\sin2\beta}=\frac 41 \\ \Rightarrow \frac{\sin2\alpha+\sin2\beta}{\sin2\alpha-\sin2\beta}=\frac{4+1}{4-1} \,\,[*] \\ \Rightarrow \frac{2\sin(\alpha+\beta)\cos(\alpha-\beta)}{2\sin(\alpha-\beta)\cos(\alpha+\beta)}=\frac 53 \\ \Rightarrow \frac{\tan(\alpha+\beta)}{\tan(\alpha-\beta)}=\frac 53 \\ \therefore 5\tan(\alpha-\beta)=3\tan(\alpha+\beta)\,\,\text{(proved)}$

Note[*] : By  componendo and dividendo

$10(i)\,\,$ If $\,\,\sin A=m\sin B,\,$ prove that, $\,\,\tan\frac{A-B}{2}=\frac{m-1}{m+1} \tan\frac{A+B}{2}.$

Sol. $\,\,\sin A=m\sin B \\ \Rightarrow \frac{\sin A}{\sin B}=\frac{m}{1} \\ \Rightarrow\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{m-1}{m+1} \\ \Rightarrow \frac{2\sin\frac{A-B}{2}\cos\frac{A+B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}=\frac{m-1}{m+1} \\ \Rightarrow \tan\frac{A-B}{2}\cot\frac{A+B}{2}=\frac{m-1}{m+1} \\ \Rightarrow  \tan\frac{A-B}{2}=\frac{m-1}{m+1} \tan\frac{A+B}{2}\,\,\text{(proved)}$

$10(ii)\,\,$ If $\,\cos\alpha=k\cos\beta,\,\,$ show that , $\,\,\tan\frac{\alpha+\beta}{2}=\frac{1-k}{1+k}\cot\frac{\alpha-\beta}{2}$

Sol.  $\,\cos\alpha=k\cos\beta \\ \Rightarrow \frac{\cos\beta}{\cos\alpha}=\frac 1k \\ \Rightarrow \frac{\cos\beta-\cos\alpha}{\cos\beta+\cos\alpha}=\frac{1-k}{1+k} \\ \Rightarrow \frac{2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}=\frac{1-k}{1+k} \\ \Rightarrow \tan\frac{\alpha+\beta}{2}\tan\frac{\alpha-\beta}{2}=\frac{1-k}{1+k} \\ \therefore \,\,\tan\frac{\alpha+\beta}{2}=\frac{1-k}{1+k}\cot\frac{\alpha-\beta}{2}\,\,\text{(proved)}$ 

$11.\,\,$ If $\,\,\csc A+\sec A=\csc B+\sec B,\,$ prove that, $\,\,\tan\frac{A+B}{2}=\cot A\cot B$

By $\csc A\,$ we mean cosec $\,A$

Sol.  $\,\,\csc A+\sec A=\csc B+\sec B\\ \Rightarrow \csc A-\csc B=\sec B-\sec A \\ \Rightarrow \frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A} \\ \Rightarrow \frac{\sin B-\sin A}{\sin A\sin B}=\frac{\cos A-\cos B}{\cos A\cos B} \\ \Rightarrow \frac{2\sin\frac{B-A}{2}\cos\frac{A+B}{2}}{\sin A\sin B}=\frac{2\sin\frac{A+B}{2}\sin\frac{B-A}{2}}{\cos A\cos B} \\ \Rightarrow \frac{\sin\frac{A+B}{2}}{\cos\frac{A+B}{2}}=\frac{\cos A\cos B}{\sin A\sin B} \\ \therefore \tan\frac{A+B}{2}=\cot A\cot B\,\,\text{(proved)}$

$12.\,\,$ If $\,\,\alpha,\beta,\gamma\,\,$ are in A.P. , show that $\,\,\cot\beta=\frac{\sin\alpha-\sin\gamma}{\cos\gamma-\cos\alpha}.$

Sol. Since $\,\,\alpha,\beta,\gamma\,\,$ are in A.P., $\alpha+\gamma=2\beta \rightarrow(1)$

Now,  $\frac{\sin\alpha-\sin\gamma}{\cos\gamma-\cos\alpha}\\=\frac{2\sin\frac{\alpha-\gamma}{2}\cos\frac{\alpha+\gamma}{2}}{2\sin\frac{\alpha+\gamma}{2}\sin\frac{\alpha-\gamma}{2}}\\=\cot\frac{\alpha+\gamma}{2}\\=\cot\beta \,\,\,[\text{By (1)}]$

$13.\,\,$ If $\,\,\sin x+\sin y=\sqrt3(\cos y-\cos x),\,\,$ then show that $\,\,\sin3x+\sin3y=0.$

Sol. $\,\,\sin x+\sin y=\sqrt3(\cos y-\cos x)\\ \Rightarrow 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}=\sqrt3 \times 2\sin\frac{x+y}{2}\sin\frac{x-y}{2} \\ \Rightarrow 2\sin\frac{x+y}{2}\left(\cos\frac{x-y}{2}-\sqrt3\sin\frac{x-y}{2}\right)=0 \\ \therefore \sin\frac{x+y}{2}=0 \,\,\text{or}\,\,\cos\frac{x-y}{2}=\sqrt3\sin\frac{x-y}{2} \\ \therefore \sin\frac{x+y}{2}=0,\,\,\text{or,}\,\,\tan\frac{x-y}{2}=\frac{1}{\sqrt3} \\ \therefore \frac{x+y}{2}=0^{\circ}\,\,\text{or,}\,\,\frac{x-y}{2}=30^{\circ}$

Now, for $\,\,\frac{x+y}{2}=0^{\circ},\\ \sin3x+\sin3y\\=2\sin\frac{3(x+y)}{2}\cos\frac{3(x-y)}{2}\\=2\sin\left(3\times 0^{\circ}\right) \cos\frac{3(x-y)}{2}\\=0$

Again, for $\,\,\frac{x-y}{2}=30^{\circ},\\ \sin3x+\sin3y\\=2\sin\frac{3(x+y)}{2}\cos\frac{3(x-y)}{2}\\=2\sin\frac 32\left(x+y\right) \cos(3 \times 30^{\circ})\\=0\,\,[\text{Since,}\,\cos90^{\circ}=0]$

If you want to download Full solution PDF of Chhaya  math solution of Transformation of Sums and Products for class XI, click here .

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