Differentiation (Part-38) | S N De

$\,\,5.\,\,$ If $\,\sin\theta=\frac 35,\,\,\tan\phi=\frac{8}{15}\,\,$ and $\,\theta\,\,$ and $\,\,\phi\,$ lie in the first and third quadrant respectively, find $\,\csc(\theta+\phi).$

Sol. $\,\,\sin\theta=\frac 35 \\ \Rightarrow \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{25}}=\frac 45.\\ \therefore \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{3/5}{4/5}=\frac 34.\\ \therefore\tan(\theta+\phi)\\=\frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}\\=\frac{\frac 34+\frac{8}{15}}{1-\frac 34. \frac{8}{15}}\\=\frac{\frac{45+32}{60}}{1-\frac 25}\\=\frac{77/60}{3/5}\\=\frac{77}{36} \\ \Rightarrow \cot(\theta+\phi)=\frac{36}{77} \\ \therefore \csc^2(\theta+\phi)\\=1+\cot^2(\theta+\phi)\\=1+(\frac{36}{77})^2\\=\frac{77^2+36^2}{77^2}\\=\frac{7225}{77^2}\\ \Rightarrow \csc(\theta+\phi)=-\frac{85}{77}\rightarrow(1)$

In $\,(1)\,$, we see that negative sign has been taken as value of $\,\,\csc(\theta+\phi)\,$ is negative in third and fourth quadrants. 

$\,\,6.\,$ Prove that, $\,(i)\,\frac{\cos20^{\circ}+\sin20^{\circ}}{\cos20^{\circ}-\sin20^{\circ}}=\tan 65^{\circ}$

Sol.  L.H.S.$=\frac{\cos20^{\circ}+\sin20^{\circ}}{\cos20^{\circ}-\sin20^{\circ}}\\=\frac{1+\tan20^{\circ}}{1-\tan20^{\circ}}\,\,[**]\\=\frac{\tan45^{\circ}+\tan20^{\circ}}{1-\tan45^{\circ}.\tan20^{\circ}}\\=\tan(45^{\circ}+20^{\circ})\\=\tan65^{\circ}\\=\text{R.H.S.}\,\,\text{(proved)}$

Note [**] : Dividing numerator and denominator by $\,\cos20^{\circ}.$

$\,\,(ii)\,\,\csc(\theta+\phi)=\frac{\csc\theta\csc\phi}{\cot\theta+\cot\phi}$

Sol. $\,\,\csc(\theta+\phi)\\=\frac{1}{\sin(\theta+\phi)}\\=\frac{1}{\sin\theta\cos\phi+\cos\theta\sin\phi}\\=\frac{1}{\sin\theta\sin\phi(\frac{\cos\phi}{\sin\phi}+\frac{\cos\theta}{\sin\theta})}\\=\frac{1}{\sin\theta\sin\phi(\cot\phi+\cot\theta)}\\=\frac{\csc\theta\csc\phi}{\cot\theta+\cot\phi}\quad \text{(proved)}$

Note : By $\csc\phi\,\,$ we mean $\,\text{cosec}\,\phi.$

$\,(iii)\,\,\cos67^{\circ}24'\cos7^{\circ}24'+\cos82^{\circ}36'\cos22^{\circ}36'=\frac 12.$

Sol. $\,\,\cos67^{\circ}24'\cos7^{\circ}24'+\cos82^{\circ}36'\cos22^{\circ}36'\\=\,\,\cos67^{\circ}24'\cos7^{\circ}24'\\+\sin(90^{\circ}-82^{\circ}36')\sin(90^{\circ}-22^{\circ}36')\\=\cos67^{\circ}24'\cos7^{\circ}24'+\sin7^{\circ}24'\sin67^{\circ}24'\\=\cos(67^{\circ}24'-7^{\circ}24')\\=\cos 60^{\circ}\\=\frac 12\quad \text{(proved)}$

$\,(iv)\,\,\tan(B-C)+\tan(C-A)+\tan(A-B)\\=\tan(B-C)\tan(C-A)\tan(A-B)$

Sol. Let $\,x=(B-C),\,\,y=(C-A),\,\,z=(A-B)$ 

so that $\,x+y+z=0 \\ \Rightarrow \tan(x+y+z)=\tan 0 \\ \Rightarrow \frac{\tan x+\tan y+\tan z-\tan x \tan y\tan z}{1-\tan x\tan y-\tan y\tan z-\tan z\tan x}=0 \\ \Rightarrow \tan x+\tan y+\tan z\\-\tan x\tan y\tan z=0 \\ \Rightarrow \tan x+\tan y+\tan z=\tan x\tan y\tan z \\ \Rightarrow \tan(B-C)+\tan(C-A)+\tan(A-B)\\=\tan(B-C)\tan(C-A)\tan(A-B)\\ \text{(proved)} $

$\,7.\,$ Show that, the value of each of the following expressions is independent of $\,\theta : $

$\,(i)\,\left[\cot(\frac{\pi}{4}-\theta)-1\right](\cot\theta-1)\\=\left[\frac{\cot\frac{\pi}{4}\cot\theta+1}{\cot\theta-\cot\frac{\pi}{4}}-1\right](\cot\theta-1)\\=\left[\frac{\cot\theta+1}{\cot\theta-1}-1\right](\cot\theta-1)\\=\frac{\cot\theta+1-\cot\theta+1}{(\cot\theta-1)} \times (\cot\theta-1)\\=2,\,\text{independent of}\,\,\,\theta.$

$\,(ii)\,\,\cos^2\left(\frac{\pi}{4}+\theta\right)-\sin^2\left(\frac{\pi}{4}-\theta\right)\\=\cos[(\frac{\pi}{4}+\theta)+(\frac{\pi}{4}-\theta)]\\ \times\cos[(\frac{\pi}{4}+\theta)-(\frac{\pi}{4}-\theta)]\\=\cos\frac{\pi}{2}\times \cos 2\theta\\=0\times \cos2\theta\\=0,\,\text{independent of}\,\,\,\theta.$

$\,8.\,$ If $\,\,\cos(\alpha-\beta)+1=0,\,\,$ show that , $\,\cos\alpha+\cos\beta=0\,\,$ and $\,\,\sin\alpha+\sin\beta=0.$

Sol. $\,\,\cos(\alpha-\beta)+1=0 \\ \Rightarrow \cos(\alpha-\beta)=-1=\cos\pi \\ \Rightarrow \alpha-\beta=\pi\rightarrow(1)$

Now, $\,\,\cos\alpha+\cos\beta\\=\cos(\pi+\beta)+\cos\beta\\=-\cos\beta+\cos\beta\,\,\,\,[\text{By (1)}]\\=0\,\,\text{(showed)}$

Again, $\,\,\sin\alpha+\sin\beta\\=\sin(\pi+\beta)+\sin\beta\,\,\,\,[\text{By (1)}]\\=-\sin\beta+\sin\beta\\=0\,\,\,\text{(showed)}$