Differentiation (Part-38) | S N De

 

$\,1(i).\,$ Prove that, $ \,\,\,\sin^2\alpha+\sin^2(120^{\circ}-\alpha)+\sin^2(120^{\circ}+\alpha)\\=\frac 32.$

Sol. $ \,\,\,\sin^2\alpha+\sin^2(120^{\circ}-\alpha)+\sin^2(120^{\circ}+\alpha)\\=\sin^2\alpha+(\sin120^{\circ}\cos\alpha-\cos120^{\circ}\sin\alpha)^2\\+(\sin120^{\circ}\cos\alpha+\cos120^{\circ}\sin\alpha)^2\\=\sin^2\alpha+2[(\sin120^{\circ}\cos\alpha)^2\\+(\cos120^{\circ}\sin\alpha)^2]\,\,\,[**]\\=\sin^2\alpha+2[\sin^260^{\circ}\cos^2\alpha\\+\cos^260^{\circ}\sin^2\alpha]\,\,[*]\\=\sin^2\alpha+2\left[\left(\frac{\sqrt3}{2}\right)^2\cos^2\alpha+\left(\frac 12\right)^2\sin^2\alpha\right]\\=\sin^2\alpha+2\left(\frac 34\cos^2\alpha+\frac 14\sin^2\alpha\right)\\=\sin^2\alpha+\frac 32\cos^2\alpha+\frac 12\sin^2\alpha\\=\frac 32\sin^2\alpha+\frac 32\cos^2\alpha\\=\frac 32(\sin^2\alpha+\cos^2\alpha)\\=\frac 32\times 1\\=\frac 32\quad\text{(proved)}$

Note [**] : $\,\,(a-b)^2+(a+b)^2=2(a^2+b^2)$

Note[*] : $\,\,\sin120^{\circ}=\sin(180^{\circ}-60^{\circ})=\sin60^{\circ}\\ \cos120^{\circ}=\cos(180^{\circ}-60^{\circ})=-\cos60^{\circ}$

$\,1(ii)\,$ Prove that, $\,\,\sin^2\left(\frac{\pi}{8}+\frac{\theta}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{\theta}{2}\right)=\frac{1}{\sqrt2}\sin\theta$

Sol. $\,\,\sin^2\left(\frac{\pi}{8}+\frac{\theta}{2}\right)-\sin^2\left(\frac{\pi}{8}-\frac{\theta}{2}\right)\\=\sin\left[\left(\frac{\pi}{8}+\frac{\theta}{2}\right)+\left(\frac{\pi}{8}-\frac{\theta}{2}\right)\right]\\ \times \sin\left[\left(\frac{\pi}{8}+\frac{\theta}{2}\right)-\left(\frac{\pi}{8}-\frac{\theta}{2}\right)\right]\\=\sin\frac{\pi}{4}\sin\theta\\=\frac{1}{\sqrt2}\sin\theta\quad\text{(proved)}$

$\,1(iii)\,\,\,\cos^2A+\cos^2\left(A+\frac{\pi}{3}\right)+\cos^2\left(A-\frac{\pi}{3}\right)\\=\frac 32.$ 

Sol. $\,\,\,\cos^2A+\cos^2\left(A+\frac{\pi}{3}\right)+\cos^2\left(A-\frac{\pi}{3}\right)\\=\cos^2A+\left(\cos A\cos \frac{\pi}{3}-\sin A\sin\frac{\pi}{3}\right)^2\\+\left(\cos A\cos\frac{\pi}{3}+\sin A\sin\frac{\pi}{3}\right)^2\\=\cos^2A+2\left[\left(\cos A\cos \frac{\pi}{3}\right)^2\\+\left(\sin A \sin \frac{\pi}{3}\right)^2\right]\,\,[*]\\=\cos^2A+2\left[\cos^2A.\left(\frac 12\right)^2+\sin^2A.\left(\frac{\sqrt3}{2}\right)^2\right]\\=\cos^2A+\frac 12\cos^2A+\frac 32\sin^2A\\=\frac 32\cos^2A+\frac 32\sin^2A\\=\frac 32\left(\sin^2A+\cos^2A\right)\\=\frac 32\quad\text{(proved)}$ 

Note [*] : $\,\,(a-b)^2+(a+b)^2=2(a^2+b^2)$

$\,1(iv)\,\,$ Prove that, $\,\,\tan 70^{\circ}=2\tan 50^{\circ}+\tan20^{\circ}.$

Sol. We have, $\,\,\tan 70^{\circ}=\tan(50^{\circ}+20^{\circ})\\ \Rightarrow \tan 70^{\circ}=\frac{\tan 50^{\circ}+\tan 20^{\circ}}{1-\tan 50^{\circ}\tan 20^{\circ}}\\ \Rightarrow \tan 70^{\circ}-\tan 20^{\circ}\tan 50^{\circ}\tan 70^{\circ}\\=\tan 50^{\circ}+\tan 20^{\circ} \\ \Rightarrow \tan 70^{\circ}-\cot(90^{\circ}-20^{\circ})\tan 70^{\circ}\tan 50^{\circ}\\=\tan 50^{\circ}+\tan 20^{\circ}\\ \Rightarrow \tan 70^{\circ}-\cot 70^{\circ}\tan 70^{\circ}\tan 50^{\circ}\\=\tan 50^{\circ}+\tan 20^{\circ} \\ \Rightarrow \tan 70^{\circ}-1\times \tan 50^{\circ}=\tan 50^{\circ}+\tan 20^{\circ} \\ \Rightarrow \tan 70^{\circ}=2\tan 50^{\circ}+\tan 20^{\circ}\quad \text{(proved)}$

$\,2.\,$ Simplify : $\,(i)\,\,\cos A\sin (B-C)+\cos B\sin (C-A)\\+\cos C \sin (A-B)$

Sol. $\,\cos A\sin (B-C)+\cos B\sin (C-A)\\+\cos C \sin (A-B)\\=\cos A(\sin B\cos C-\cos B\sin C)\\+\cos B(\sin C \cos A-\cos C\sin A)\\+\cos C(\sin A\cos B-\cos A \sin B)\\=\cos A\sin B\cos C-\cos B\sin C\cos A\\+\cos B\sin C \cos A-\cos C\sin A \cos B\\+\cos C\sin A\cos B-\cos A \sin B \cos C\\=0$

$\,2.\,$ Simplify : $\,(ii)\,\,1+\frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}$

Sol. $\,\,\frac{\sin(A-B)}{\cos A\cos B}\\=\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}\\=\tan A-\tan B\rightarrow(1)$

Similarly,  $\,\,\frac{\sin(B-C)}{\cos B\cos C}\\=\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}\\=\tan B-\tan C\rightarrow(2)$

and $\,\,\frac{\sin(C-A)}{\cos C\cos A}\\=\frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A}\\=\tan C-\tan A\rightarrow(3)$

So, $\,1+\frac{\sin(A-B)}{\cos A\cos B}+\frac{\sin(B-C)}{\cos B\cos C}+\frac{\sin(C-A)}{\cos C\cos A}\\=1+\tan A-\tan B+\tan B-\tan C\\+\tan C-\tan A\,\,\,[\text{By (1),(2),(3)}]\\=1$

$\,2.\,$ Simplify : $\,(iii)\,\,\tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{3\pi}{4}+\theta\right)$

Sol. $\,\,\tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{3\pi}{4}+\theta\right)\\=\tan\left(\frac{\pi}{4}+\theta\right)\tan\left[\pi-(\frac{\pi}{4}-\theta)\right]\\=\tan\left(\frac{\pi}{4}+\theta\right).\left[-\tan\left(\frac{\pi}{4}-\theta\right)\right]\\=\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta} \times \left[-\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\right]\\=-\frac{1+\tan\theta}{1-\tan\theta} \times \frac{1-\tan\theta}{1+\tan\theta}\\=-1\quad\text{(Ans.)}$

$\,2.\,$ Simplify : $\,(iv)\,\,\sin(B+C)\sin(B-C)+\sin(C+A)\\\times \sin(C-A)+\sin(A+B)\sin(A-B)$

Sol. $\,\,\sin(B+C)\sin(B-C)+\sin(C+A)\\\times \sin(C-A)+\sin(A+B)\sin(A-B)\\=\sin^2B-\sin^2C+\sin^2C-\sin^2A\\+\sin^2A-\sin^2B\\=0\quad\text{(ans.)}$