Ad-1

if( aicp_can_see_ads() ) {

TRANSFORMATIONS OF SUMS AND PRODUCTS (Part-2)

 

TRANSFORMATIONS OF SUMS AND PRODUCTS (Part-2)


$\,6.\,$ If $\,\,A+B+C=\pi,\,\,\sin(A+\frac C2)=n\sin\frac C2,\,\,$ show that, $\,\tan\frac A2\tan \frac B2=\frac{n-1}{n+1}.$

Sol. We have, $\,\,\,\sin(A+\frac C2)=n\sin\frac C2 \cdots(1),\\ A+B+C=\pi \cdots(2)$

Now,$\,\,\frac{n-1}{n+1}\\=\frac{\frac{\sin\left(A+\frac C2\right)}{\sin\frac C2}-1}{\frac{\sin\left(A+\frac C2\right)}{\sin\frac C2}+1}\\=\frac{\sin\left(A+\frac C2\right)-\sin\frac C2}{\sin\left(A+\frac C2\right)+\sin \frac C2}\\=\frac{2\sin\frac A2\cos\frac{A+C}{2}}{2\sin\frac{A+C}{2}\cos \frac A2}\\=\tan\frac A2\cot\left(\frac{\pi}{2}-\frac B2\right)\,[\text{By (2)}]\\ \therefore \tan\frac A2\tan\frac B2=\frac{n-1}{n+1}$

$\,7.\,$ If $\,\,\sin\theta+\sin\phi=a,\,\cos\theta+\cos\phi=b,\,\,$ prove that, $\,\tan\frac{\theta-\phi}{2}=\pm\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$

Sol.  $\,\,a=\sin\theta+\sin\phi=2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2},\\ b=\cos\theta+\cos\phi=2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}. \\ \therefore a^2+b^2\\=4\cos^2\frac{\theta-\phi}{2}\left(\sin^2\frac{\theta+\phi}{2}+\cos^2\frac{\theta+\phi}{2}\right)\\=4\cos^2\frac{\theta-\phi}{2} \\ \therefore \tan^2\frac{\theta-\phi}{2}\\=\frac{1-\frac{a^2+b^2}{4}}{\frac{a^2+b^2}{4}}\\=\frac{4-a^2-b^2}{a^2+b^2}$

$\,8.\,$ Prove that, $\,\,\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n=2\cot^n\frac{A-B}{2}\,\,\text{or}\,\,0\,$ accordingly as $\,n\,$ is an even or odd integer.

Sol. Let $\,\,P=\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n\\=\left(\frac{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}{2\sin\frac{A-B}{2}\cos\frac{A+B}{2}}\right)^n+\left(\frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\sin\frac{A+B}{2}\sin\frac{B-A}{2}}\right)^n\\=\left[\cot\frac{A-B}{2}\right]^n+\left[-\cot\frac{A-B}{2}\right]^n \cdots(1)$

Hence, from (1), we can conclude that $\,P=\cot^n\frac{A-B}{2}+\cot^n\frac{A-B}{2}\\~~~~=2\cot^n\frac{A-B}{2}\,\,[\text{For n=even}] \\ \text{and}\,\,P=\cot^n\frac{A-B}{2}-\cot^n\frac{A-B}{2}\\~~~~~~~~~~~=0\,\, \text{[For n=odd]}$

$\,9.\,$ If $\,\,\sin A=\frac{1}{\sqrt2},\,\,\sin B=\frac{1}{\sqrt3},\,\,$ find the value of $\,\tan\frac{A+B}{2}\cot\frac{A-B}{2}.$

Sol. $\,\tan\frac{A+B}{2}\cot\frac{A-B}{2}\\=\frac{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}\\=\frac{\sin A+\sin B}{\sin A-\sin B}\\=\frac{\frac{1}{\sqrt2}+\frac{1}{\sqrt3}}{\frac{1}{\sqrt2}-\frac{1}{\sqrt3}}\\=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\\=\frac{(\sqrt3+\sqrt2)(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}\\=\frac{(\sqrt3+\sqrt2)^2}{(\sqrt3)^2-(\sqrt2)^2}\\=\frac{3+2+2\sqrt6}{3-2}\\=5+2\sqrt6$

$\,10.\,$ Find the value of $\,A\,$ for which $\,\,\cos A\sin (A-\frac{\pi}{6})\,$ is maximum.

Sol. $\,\,\cos A\sin (A-\frac{\pi}{6})\\=\frac 12\left[2\cos A\sin (A-\pi/6)\right]\\=\frac 12\left[\sin(2A-\pi/6)-\sin \frac{\pi}{6}\right]\cdots(1)$

Now, the expression given in (1) will be maximum when $\,\,\sin(2A-\pi/6)=1 \\ \Rightarrow 2A-\pi/6=\pi/2 \\ \Rightarrow A=\frac{\pi}{3}$

$\,11.\,$ The two different values of $\,\theta,\,$ namely $\,\theta_1\,\text{and}\,\,\theta_2(0 \leq \theta_1<2\pi,\,\,\theta_2(0 \leq \theta_2<2\pi)\,\,$ satisfy the equation $\,\,\sin(\theta+\phi)=\frac 12\sin2\phi.$ 

Prove that $\,\,\frac{\sin\theta_1+\sin \theta_2}{\cos\theta_1+\cos \theta_2}=\cot \phi.$

Sol. Since the two different values of $\,\theta,\,$ namely $\,\theta_1\,\text{and}\,\,\theta_2(0 \leq \theta_1<2\pi,\,\,\theta_2(0 \leq \theta_2<2\pi)\,\,$ satisfy the equation $\,\,\sin(\theta+\phi)=\frac 12\sin2\phi, \\ \sin(\theta_1+\phi)=\frac 12\sin2\phi \\ \text{and}\,\sin(\theta_2+\phi)=\frac 12 \sin2\phi \\ \therefore \sin(\theta_1+\phi)=\sin(\theta_2+\phi) \\ \Rightarrow \theta_1+\phi=\pi-\theta_2-\phi \,\,[\theta_1 \neq \theta_2] \\ \Rightarrow \theta_1+\theta_2=\pi- 2\phi \\ \text{Now,}\,\,\frac{\sin\theta_1+\sin\theta_2}{\cos\theta_1+\cos\theta_2}\\=\frac{2\sin\frac{\theta_1+\theta_2}{2}\cos\frac{\theta_1-\theta_2}{2}}{2\cos\frac{\theta_1+\theta_2}{2}\cos\frac{\theta_1-\theta_2}{2}}\\=\tan\frac{\theta_1+\theta_2}{2}\\=\tan\frac{\pi-2\phi}{2}\\=\tan\left(\pi/2-\phi\right)\\=\cot\phi$ 

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


Post a Comment

0 Comments
* Please Don't Spam Here. All the Comments are Reviewed by Admin.