Differentiation (Part-38) | S N De

 

$\,2.\,$ Prove: 

$\,\,(i)\,\,\frac{(\sin 0^{\circ}+\sin 60^{\circ})(\cos 60^{\circ}+\cot 45^{\circ})}{(\cot 60^{\circ}+\tan 30^{\circ})(\csc 30^{\circ}-\csc 90^{\circ})}=\frac 98$

Sol. $\quad \frac{(\sin 0^{\circ}+\sin 60^{\circ})(\cos 60^{\circ}+\cot 45^{\circ})}{(\cot 60^{\circ}+\tan 30^{\circ})(\csc 30^{\circ}-\csc 90^{\circ})}\\=\frac{\left(0+\frac{\sqrt3}{2}\right)\left(\frac 12+1\right)}{\left(\frac{1}{\sqrt3}+\frac{1}{\sqrt3}\right)\left(2-1\right)}\\=\frac{\frac{\sqrt3}{2} \times \frac 32}{\frac{2}{\sqrt3} \times 1}\\=\frac{\frac{3\sqrt3}{4}}{\frac{2}{\sqrt3}}\\=\frac{3\sqrt3}{4} \times \frac{\sqrt3}{2}\\=\frac 98\quad \text{(proved)}$

Note : Here, by $\,\,\csc \theta \,$ we mean $\text{cosec}\,\theta. $

$\,\,(ii)\,\,\cos^260 ^{\circ}, \cos^245 ^{\circ},\cos^230 ^{\circ}\,\,$ are in A.P.

Sol. Let $\,\,p=\cos^260 ^{\circ}, \,\,q=\cos^245 ^{\circ},\,\,r=\cos^230 ^{\circ}$

Now, $\,\,p=\cos^260 ^{\circ}=(\frac 12)^2=\frac 14 ,\\ q=\cos^245 ^{\circ}=(\frac{1}{\sqrt2})^2=\frac 12, \\ r=\cos^230^{\circ}=(\frac{\sqrt3}{2})^2=\frac 34. \\ \therefore p+r=\frac 14+\frac 34=1=2q \\ \Rightarrow r-q=q-p \rightarrow(1)$

Hence, from $\,\,(1),\,\,$ it follows that $\quad\cos^260 ^{\circ}, \cos^245 ^{\circ},\cos^230 ^{\circ}\,\,$ are in A.P.

$\,\,(iii)\,\,\tan\frac{\pi}{3},\,\,\tan\frac{\pi}{4},\,\,\tan\frac{\pi}{6}\,\,$ are in G.P.

Sol. Let $\,\,p=\tan\frac{\pi}{3},\,\,q=\tan\frac{\pi}{4},\,\,r=\tan\frac{\pi}{6}\,\,$ so that $\,\,p=\sqrt3,q=1,r=\frac{1}{\sqrt3} \rightarrow(1)$

From $\,\,(1),\,\,$ it follows that $\,\,pr=\sqrt3 \times \frac{1}{\sqrt3}=1=q^2 \\ \Rightarrow \frac pq=\frac qr\rightarrow(2).$

Hence, from $\,\,(2),\,\,$ we can conclude that  $\,\,\tan\frac{\pi}{3},\,\,\tan\frac{\pi}{4},\,\,\tan\frac{\pi}{6}\,\,$ are in G.P.

Read More :  TRIGONOMETRIC RATIOS OF POSITIVE ACUTE ANGLES (PART-10)

$\,\,(iv)\,\, 2\cos^2\frac{\pi}{3}+\frac 34\sec^2\frac{\pi}{4}+4\sin^2\frac{\pi}{6}=\cot^2\frac{\pi}{6}.$

Sol. $\,\, 2\cos^2\frac{\pi}{3}+\frac 34\sec^2\frac{\pi}{4}+4\sin^2\frac{\pi}{6}\\=2(\frac 12)^2+\frac 34\times (\sqrt2)^2+4\times (\frac 12)^2\\=2\times \frac 14+\frac 34 \times 2+4 \times \frac 14\\=\frac 12+\frac 32+1\\=\frac{1+3}{2}+1\\=\frac 42+1\\=2+1\\=3\\=(\sqrt3)^2\\=\left(\cot\frac{\pi}{6}\right)^2\\=\cot^2\frac{\pi}{6}\quad \text{(proved)}$

$\,\,3.\,\,$ The triangle  $\,A\,$ of $\,\,\Delta ABC\,\,$ is obtuse ; if $\,\,\sec(B+C)=\csc(B-C)=2,\,\,$ find the angles. 

Sol. We have , $\,\,\sec(B+C)=2=\sec 60^{\circ} \\ \Rightarrow B+C=60^{\circ}\rightarrow(1) \\ \text{Again,}\,\,\csc(B-C)=2=\csc 30^{\circ} \\ ~~~~~~~~\Rightarrow  B-C=30  ^{\circ}\rightarrow(2)$

From $\,\,(1),(2)\,\,$ we get, $\,\,(B+C)+(B-C)=60 ^{\circ}+30 ^{\circ} \\ \Rightarrow 2B=90  ^{\circ} \\ \Rightarrow B=\frac{90^{\circ}}{2}=45^{\circ}$

Again, By $\,\,(1)\,\,$ we get, $\,\, 45^{\circ}+C=60^{\circ} \\ \Rightarrow C=60^{\circ}-45^{\circ}\\~~~~~~~~=15^{\circ}.$

Hence, $\,\,A=180^{\circ}-(B+C)=180^{\circ}-60^{\circ}=120^{\circ} \\ \therefore A=120^{\circ},\,\,B=45^{\circ},\,\,C=15^{\circ}.$

Note : Here, by $\,\,\csc \theta \,$ we mean $\text{cosec}\,\theta. $

$\,\,4.\,\,$ If $\,\,\theta\,\,$ is a positive acute angle and $\,\,(i)\,\,\tan\theta=\frac 34,\,\,$ find $\,\,\sin\theta,\,\,\sec\theta.$

Sol. $\,\,\tan\theta=\frac 34\,\,\text{(given)} \\ \Rightarrow \tan^2\theta=\left(\frac 34\right)^2 \\ \Rightarrow 1+\tan^2\theta=1+\frac{9}{16} \\ \Rightarrow \sec^2\theta=\frac{16+9}{16}\\ \Rightarrow \sec\theta=\sqrt{\frac{25}{16}}=\frac 54 \rightarrow (1) \\ \Rightarrow \cos\theta=1/\sec\theta=\frac 45 \\ \therefore \sin\theta=\sqrt{1-\cos^2\theta}\\~~~~~~~~~~~=\sqrt{1-\left(\frac 45\right)^2}\\~~~~~~~~~~~=\sqrt{1-\frac{16}{25}}\\~~~~~~~~~~~=\sqrt{\frac{25-16}{25}}\\~~~~~~~~~~~=\sqrt{\frac{9}{25}}\\~~~~~~~~~~~=\frac 35\rightarrow(2)$

From $\,\,(1),(2)\,\,$ we get the required values of $\,\,\sin\theta,\,\,\sec\theta.$

$\,\,4.\,\,$ If $\,\,\theta\,\,$ is a positive acute angle and $\,\,(ii)\,\,\sec\theta=\frac{m^2+1}{2m},\,\,$ find $\,\,\tan\theta,\,\,\csc\theta.$

Sol. We have, $\,\,\sec\theta=\frac{m^2+1}{2m} \rightarrow(1)\\ \Rightarrow \sec^2\theta=\left(\frac{m^2+1}{2m}\right)^2 \\ \Rightarrow 1+\tan^2\theta=\left(\frac{m^2+1}{2m}\right)^2 \\ \Rightarrow \tan^2\theta=\left(\frac{m^2+1}{2m}\right)^2-1\\~~~~~~~~~~~~~~~=\frac{(m^2+1)^2-4m^2}{4m^2}\\~~~~~~~~~~~~~~~=\frac{(m^2-1)^2}{4m^2} \\ \Rightarrow \tan\theta=\sqrt{\frac{(m^2-1)^2}{4m^2}}=\frac{m^2-1}{2m}\rightarrow(2)$ 

Now, using $\,(1)\,\,$ we get, $\,\,\cos\theta=\frac{2m}{m^2+1} \\ \Rightarrow \sin\theta=\sqrt{1-\cos^2\theta}\\~~~~~~~~~~~~=\sqrt{1-\left(\frac{2m}{m^2+1}\right)^2}\\~~~~~~~~~~~~=\sqrt{\frac{(m^2+1)^2-4m^2}{(m^2+1)^2}}\\~~~~~~~~~~~~=\sqrt{\frac{(m^2-1)^2}{(m^2+1)^2}}\\~~~~~~~~~~~~=\frac{m^2-1}{m^2+1} \\ \therefore \csc \theta=\frac{m^2+1}{m^2-1}\rightarrow(3)$

From $\,\,(2),(3)\,\,$ we get the required values of $\,\,\tan\theta,\,\,\csc\theta.$

Note : Here, by $\,\,\csc \theta \,$ we mean $\text{cosec}\,\theta. $

$\,\,4.\,\,$ If $\,\,\theta\,\,$ is a positive acute angle and $\,\,(iii)\,\,\cos\theta=\frac{15}{17},\,\,$ find the values of other trigonometric ratios. 

Sol.  Let us consider a right-angled triangle whose base $\,(b)\,=15k,\,k(\neq0)\,\,$ being a constant and hypotenuse $(h)=17k,\,\,$ so that $\,\,\cos\theta=\frac{15}{17}=\frac bh.$ 

For a right-angled triangle, we know $\,\,h^2=p^2+b^2,\,\,[\text{p=perpendicular}]\\ \Rightarrow (17k)^2=p^2+(15k)^2 \\ \Rightarrow p=k\sqrt{17^2-15^2}\\~~~~~~~=k\sqrt{64}\\~~~~~~~=8k.$

Now, $\,\,\sin\theta=\frac ph=\frac{8k}{17k}=\frac{8}{17} \\ \csc\theta=\frac hp=\frac{17k}{8k}=\frac{17}{8} \\ \tan\theta=\frac pb=\frac{8k}{15k}=\frac{8}{15} \\ \cot\theta=\frac bp=\frac{15k}{8k}=\frac{15}{8} \\ \sec\theta=\frac hb=\frac{17k}{15k}=\frac{17}{15}.$

Note : Here, by $\,\,\csc \theta \,$ we mean $\text{cosec}\,\theta. $