Differentiation (Part-38) | S N De

 $\,2.\,$ Solve $\,\,(0^{\circ } \leq \theta \leq 90^{\circ})$

$\,2(i)\,\,\,\tan\theta+\cot\theta=2 \\ \Rightarrow x+\frac 1x=2\,\,\,[\text{where}\,\,x=\tan\theta]\\ \Rightarrow x^2+1=2x \\ \Rightarrow x^2-2x+1=0 \\ \Rightarrow (x-1)^2=0 \\ \Rightarrow x=1 \\ \Rightarrow \tan\theta=1=\tan45^{\circ} \\ \Rightarrow \theta=45^{\circ}.$

$\,\,2(ii)\,\,2\cos^2\theta+5\sin\theta=4 \\ \Rightarrow 2(1-\sin^2\theta)+5\sin\theta=4 \\ \Rightarrow 2(1-x^2)+5x=4\,\,[\text{where}\,\,x=\sin\theta]\\ \Rightarrow 2-2x^2+5x=4 \\ \Rightarrow 0=2x^2-5x+4-2 \\ \Rightarrow 2x^2-5x+2=0 \\ \Rightarrow 2x^2-4x-x+2=0 \\ \Rightarrow 2x(x-2)-1(x-2)=0 \\ \Rightarrow (x-2)(2x-1)=0 \\ \Rightarrow 2x-1=0\,\,[\text{since,}\,\,x \neq 2\,\, \text{as}\,\,\sin\theta \neq2]\\ \Rightarrow x=\frac 12 \\ \Rightarrow \sin\theta=\frac 12=\sin30^{\circ}\\ \Rightarrow \theta=30^{\circ}.$

To download full PDF  of TRIGONOMETRIC RATIOS OF POSITIVE ACUTE ANGLES,  click here.

$\,2(iii)\,\,\tan^2\theta-(\sqrt3+1)\tan\theta+\sqrt3=0 \\ \Rightarrow x^2-(\sqrt3+1)x+\sqrt3=0\\~~~~~~~~~~~~~[\text{where,}\,\,x=\tan\theta]\\ \Rightarrow x^2-\sqrt3x-x+\sqrt3=0\\ \Rightarrow x(x-\sqrt3)-1(x-\sqrt3)=0\\ \Rightarrow (x-\sqrt3)(x-1)=0 \\ \Rightarrow x-\sqrt3=0,\,\,x-1=0\\ \Rightarrow x=\sqrt3,\,\,x=1\\ \text{Now,}\,\,x=\sqrt3 \Rightarrow\tan\theta=\sqrt3=\tan60^{\circ} \\ ~~~~~~~~~~~~~~~~~~~~\Rightarrow \theta=60^{\circ}$

Again, $\,\,x=1 \Rightarrow \tan\theta=1=\tan45^{\circ} \\~~~~~~~~~~~ \Rightarrow \theta=45^{\circ}$

$\,2(iv)\,\,\sec^2\theta+\tan^2\theta=7 \\ \Rightarrow (1+\tan^2\theta)+\tan^2\theta=7 \\ \Rightarrow 1+x^2+x^2=7 \,\,\,[\text{where,}\,\,x=\tan\theta]\\ \Rightarrow2x^2=7-1\\ \Rightarrow x^2=6/2=3 \\ \Rightarrow x=\sqrt3\\ \Rightarrow \tan \theta=\sqrt3=\tan60^{\circ}\\ \Rightarrow\theta=60^{\circ}.$

$\,2(v)\,\,2\sin^2\theta=3(1-\cos\theta)\\ \Rightarrow 2(1-\cos^2\theta)=3(1-\cos\theta)\\ \Rightarrow 2(1+\cos\theta)(1-\cos\theta)-3(1-\cos\theta)=0 \\ \Rightarrow (1-\cos\theta)[2(1+\cos\theta)-3]=0 \\ \Rightarrow (1-\cos\theta)(2+2\cos\theta-3)=0 \\ \Rightarrow (1-\cos\theta)(2\cos\theta-1)=0 \\ \Rightarrow 1-\cos\theta=0,\,\,2\cos\theta-1=0 \\ \Rightarrow \cos\theta=1,\,\,\cos\theta=\frac 12 \\ \text{Now,}\,\,\cos\theta=1=\cos0^{\circ} \\ ~~~~~~~~\Rightarrow \theta=0^{\circ}.$

Again, $\,\,\cos\theta=\frac 12=\cos60^{\circ} \\ \Rightarrow\theta=60^{\circ}.$

$\,\,2(vi)\,\,2\sin\theta\tan\theta+1=\tan\theta+2\sin\theta \\ \Rightarrow 2\sin\theta\tan\theta-2\sin\theta-\tan\theta+1=0 \\ \Rightarrow 2\sin\theta(\tan\theta-1)-1(\tan\theta-1)=0\\ \Rightarrow (\tan\theta-1)(2\sin\theta-1)=0 \\ \Rightarrow \tan\theta-1=0,\,\,2\sin\theta-1=0 \\ \text{Now,}\,\,\tan\theta-1=0 \\ \Rightarrow \tan\theta=1=\tan45^{\circ}\\ \Rightarrow \theta=45^{\circ}.$

Again, $\,\,2\sin\theta-1=0 \\ \Rightarrow \sin\theta=\frac 12=\sin30^{\circ}\\ \Rightarrow \theta=30^{\circ}$

$\,\,3.\,\,$ If $\,\alpha,\beta,\gamma\,\,$ are positive acute angles and $\,\,\sin(\alpha+\beta-\gamma)=\cos(\beta+\gamma-\alpha)\\=\tan(\gamma+\alpha-\beta)=1.\,\,$

 Find $\,\alpha,\beta,\gamma.$

Sol. $\,\,\sin(\alpha+\beta-\gamma)=\cos(\beta+\gamma-\alpha)=\tan(\gamma+\alpha-\beta)=1\\ \Rightarrow \sin(\alpha+\beta-\gamma)=1=\sin90^{\circ} \\ \Rightarrow \alpha+\beta-\gamma=90^{\circ}\rightarrow(1)$

Similarly, $\,\,\cos(\beta+\gamma-\alpha)=1=\cos0^{\circ}\\ \Rightarrow \beta+\gamma-\alpha=0^{\circ}\rightarrow(2)$

and $\,\,\tan(\gamma+\alpha-\beta)=1=\tan45^{\circ}\\ \Rightarrow \gamma+\alpha-\beta=45^{\circ}\rightarrow(3)$

Adding $\,(1),(2),(3)\,\,$, we get $\,\,\alpha+\beta+\gamma\\=90^{\circ}+0^{\circ}+45^{\circ}\\=135^{\circ}\rightarrow(4)$

From $\,(1)\,$ and $\,\,(4)\,$, we get, 

$(\alpha+\beta+\gamma)-(\alpha+\beta-\gamma)=135^{\circ}-90^{\circ}\\ \Rightarrow 2\gamma=45^{\circ} \\ \Rightarrow \gamma=\frac{45^{\circ}}{2}=22.5^{\circ}$

From $\,(2)\,$ and $\,\,(4)\,$, we get, 

$(\alpha+\beta+\gamma)-(\beta+\gamma-\alpha)=135^{\circ}-0^{\circ}\\ \Rightarrow 2\alpha=135^{\circ} \\ \Rightarrow \alpha=\frac{135^{\circ}}{2}=67.5^{\circ}$

From $\,(3)\,$ and $\,\,(4)\,$, we get, 

$(\alpha+\beta+\gamma)-(\gamma+\alpha-\beta)=135^{\circ}-45^{\circ}\\ \Rightarrow 2\beta=90^{\circ} \\ \Rightarrow \beta=\frac{90^{\circ}}{2}=45^{\circ}$

$\,4.\,\,$ If $\,\,A,B,C\,\,$are the angles of an acute angled triangle and $\,\,\cos(B+C-A)=0,\,\,\sin(C+A-B)=\frac{\sqrt{3}}{2},\,\,$ find the values of $\,\,A,B,C.$

Sol. If $\,\,A,B,C\,\,$are the angles of an acute angled triangle, $\,\,A+B+C=180^{\circ}\rightarrow(1)$

Again, $\,\,\cos(B+C-A)=0 =\cos90^{\circ}\\ \Rightarrow B+C-A=90^{\circ}\rightarrow(2) \\\sin(C+A-B)=\frac{\sqrt{3}}{2}=\sin60^{\circ}\\ \Rightarrow C+A-B=60^{\circ}\rightarrow(3) $

Now, from $\,(1)\,$ and $\,(2)\,\,$, we get 

$\,(A+B+C)-(B+C-A)=180^{\circ}-90^{\circ}\\ \Rightarrow 2A=90^{\circ} \\ \Rightarrow A=\frac{90^{\circ}}{2}=45^{\circ}\rightarrow(4)$

Again, from $\,(1)\,$ and $\,(3)\,\,$, we get 

$\,(A+B+C)-(C+A-B)=180^{\circ}-60^{\circ}\\ \Rightarrow 2B=120^{\circ} \\ \Rightarrow B=\frac{120^{\circ}}{2}=60^{\circ}\rightarrow(5)$

Now, from $\,\,(1),(4), (5)\,\,$, we get ,

$45^{\circ}+60^{\circ}+C=180^{\circ} \\ \Rightarrow C=180^{\circ}-(45^{\circ}+60^{\circ})\\ \Rightarrow C=180^{\circ}-105^{\circ}\\~~~~~~~~=75^{\circ}$