Differentiation (Part-38) | S N De

 $\,\,5(i)\,\,$ If $\,\,1+\cos^2A=3\cos A\sin A,\,\,$ find the value of $\,\,\cot A.$

Sol.  If $\,\,1+\cos^2A=3\cos A\sin A \\ \Rightarrow \frac{1}{\cos^2A}+1=3\frac{\sin A}{\cos A} \,\,[**]\\ \Rightarrow \sec^2A +1=3\tan A \\ \Rightarrow 1+\tan^2A+1=3\tan A \\ \Rightarrow 1+x^2+1=3x\,\,[\text{where}\,\,x=\tan A] \\ \Rightarrow x^2+2=3x \\ \Rightarrow x^2-3x+2=0 \\ \Rightarrow x^2-x-2x+2=0 \\ \Rightarrow x(x-1)-2(x-1)=0 \\ \Rightarrow (x-1)(x-2)=0 \\ \Rightarrow x-1=0,\,\,x-2=0 \\ \Rightarrow x=1,\,\,\,2 \\ \Rightarrow \tan A=1,\,\,\,\,2  \\ \Rightarrow \cot A=\frac{1}{\tan A}=1,\,\,\,\frac 12.$

Note[**] : Dividing both sides of the equation by $\,\,\cos^2A.$

$\,\,5(ii)\,\,$ If $\,\,3\sin\theta+4\cos\theta=5\,\,\,\left(0<\theta<\frac{\pi}{2}\right),\,\,$ show that $\,\,\,\sin\theta=\frac 35.$

Sol.  We have , $\,\,3\sin\theta+4\cos\theta=5\rightarrow(1)$

On squaring both sides of $\,(1)\,$, we get 

$\,\,(3\sin\theta+4\cos\theta)^2=5^2 \\ \Rightarrow (3\sin\theta)^2+(4\cos\theta)^2\\+2\times 3\sin\theta\times 4\cos\theta=25 \\ \Rightarrow 9\sin^2\theta+16\cos^2\theta+24\sin\theta\cos\theta=25 \\ \Rightarrow 9(1-\cos^2\theta)+16(1-\sin^2\theta)\\+24 \sin\theta\cos\theta=25 \\ \Rightarrow 9-9\cos^2\theta+16-16\sin^2\theta\\+24 \sin\theta\cos\theta=25 \\ \Rightarrow 25-9\cos^2\theta-16\sin^2\theta+24\sin\theta\cos\theta\\=25 \\ \Rightarrow 0=9\cos^2\theta-24\sin\theta\cos\theta+16\sin^2\theta \\ \Rightarrow (3\cos\theta)^2-2\times 3\cos\theta\times 4\sin\theta\\+(4\sin\theta)^2=0 \\ \Rightarrow (3\cos\theta-4\sin\theta)^2=0 \\ \Rightarrow 3\cos\theta-4\sin\theta=0 \\ \Rightarrow 3\cos\theta=4\sin\theta \\ \Rightarrow (3\cos\theta)^2=(4\sin\theta)^2 \\ \Rightarrow 9\cos^2\theta=16\sin^2\theta \\ \Rightarrow 9(1-\sin^2\theta)=16\sin^2\theta \\ \Rightarrow 9=16\sin^2\theta+9\sin^2\theta \\ \Rightarrow 25 \sin^2\theta=9 \\ \Rightarrow \sin^2\theta=\frac{9}{25} \\ \Rightarrow \sin\theta=\frac 35.$

$\,6.\,\,$ If $\,\,\csc\alpha+\cot\alpha=a,\,\,$ prove that, $\,\,\cos\alpha=\frac{a^2-1}{a^2+1}.$

Here, by $\,\,\csc \alpha\,$ we mean $\text{cosec}\,\alpha. $

Sol.   It is given  $\,\,\csc\alpha+\cot\alpha=a\rightarrow(1)$

We know, $\,\,\csc^2\alpha-\cot^2\alpha=1 \\ \Rightarrow (\csc\alpha+\cot\alpha)(\csc\alpha-\cot\alpha)=1 \\ \Rightarrow a(\csc\alpha-\cot\alpha)=1\,\,[\text{By (1)}] \\ \Rightarrow \csc\alpha-\cot\alpha=\frac 1a \rightarrow(2)$ 

Adding $\,\,(1) ,(2)\,\,$ we get, $\,\,2\csc \alpha=a+\frac 1a \\ \Rightarrow \csc\alpha=\frac{a^2+1}{2a} \\ \Rightarrow \sin\alpha=\frac{2a}{a^2+1}$

We know, $\,\,\sin^2\alpha+\cos^2\alpha=1 \\ \Rightarrow \left(\frac{2a}{a^2+1}\right)^2 +\cos^2\alpha=1 \\ \Rightarrow \cos^2\alpha=1-\frac{4a^2}{(a^2+1)^2}\\~~~~~~~~~~~~~~~=\frac{(a^2+1)^2-4a^2}{(a^2+1)^2}\\~~~~~~~~~~~~~~~=\frac{(a^2-1)^2}{(a^2+1)^2} \\ \therefore  \cos\alpha=\sqrt{\frac{(a^2-1)^2}{(a^2+1)^2} }=\frac{a^2-1}{a^2+1}\,\,\,\text{(proved)}$

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

$\,\,7.\,\,$ If $\,\,4x\sec A=1+4x^2,\,\,$ prove that, $\,\,\sec A+\tan A=2x\,\,$ or $\,\,\frac{1}{2x}.$

Sol. We have , $\,\,4x\sec A=1+4x^2\\ \Rightarrow  \sec A=\frac{1+4x^2}{4x} \\ \Rightarrow  \cos A=\frac{4x}{1+4x^2}\rightarrow(1)$

Now, we know $\,\,\sin^2A+\cos^2A=1 \\ \Rightarrow  \sin^2A+\left(\frac{4x}{1+4x^2}\right)^2=1 \\ \Rightarrow  \sin A= \pm\sqrt{1-\frac{16x^2}{(1+4x^2)^2}}\\~~~~~~~~~~~~~=\pm \frac{\sqrt{(1+4x^2)^2-16x^2}}{1+4x^2}\\~~~~~~~~~~~~~=\pm\frac{\sqrt{(1-4x^2)^2}}{1+4x^2}\\~~~~~~~~~~~~~=\pm\frac{1-4x^2}{1+4x^2}\rightarrow(2)$

Now, $\,\,\sec A+\tan A\\=\frac{1}{\cos A}+\frac{\sin A}{\cos A}\\=\frac{1+\sin A}{\cos A}\rightarrow(3)$

So, $\,1+\sin A\\=1 \pm \frac{1-4x^2}{1+4x^2}\\=\frac{(1+4x^2)\pm (1-4x^2)}{1+4x^2}\\=\frac{2}{1+4x^2}\,\,[\text{taking +ve value}] \rightarrow(4)\\\text{or},\,\,\frac{8x^2}{1+4x^2}\,\,[\text{taking -ve value}]\rightarrow(5)$

Now, using $\,\,(1),(3),(4)\,\,$ we get, 

$\,\,\sec A+\tan A\\=\frac{2/(1+4x^2)}{4x/(1+4x^2)}\\=\frac{2}{4x}\\=\frac{1}{2x}$

Again, using $\,\,(1),(3),(5)\,\,$ we get, 

$\,\,\sec A+\tan A\\=\frac{8x^2/(1+4x^2)}{4x/(1+4x^2)}\\=\frac{8x^2}{4x}\\=2x.$

Hence follows the result.

$\,8.\,\,$ If $\,\,\tan^2\theta=1-e^2,\,\,$ prove that, $\,\,\sec\theta+\tan^3\theta\csc\theta=(2-e^2)^{\frac 32}$

Here, by $\,\,\csc \alpha\,$ we mean $\text{cosec}\,\alpha. $

Sol. We have, $\,\,\tan^2\theta=1-e^2\\ \Rightarrow \tan\theta=(1-e^2)^{\frac 12} \\ \Rightarrow \tan\theta=(1-e^2)^{\frac 32} \rightarrow(1)$

Now, $\,\,\sec^2\theta=1+\tan^2\theta\\~~~~~~~~~~~=1+1-e^2 \\ \Rightarrow \sec\theta=(2-e^2)^{\frac 12} \rightarrow(2)$

Again, $\,\,\csc^2\theta=1+\cot^2\theta \\~~~~~~~~~~~ =1+\frac{1}{1-e^2}\\~~~~~~~~~~~ =\frac{2-e^2}{1-e^2} \\ \Rightarrow \csc\theta=\left(\frac{2-e^2}{1-e^2}\right)^{\frac 12} \rightarrow(3)$

Hence, $\,\,\sec\theta+\tan^3\theta\csc\theta\\=(2-e^2)^{\frac 12}+(1-e^2)^{\frac 32} \times \left(\frac{2-e^2}{1-e^2}\right)^{\frac 12}\\~~~~~~~[\text{By (1),(2),(3)}]\\=(2-e^2)^{\frac 12}+(1-e^2)(2-e^2)^{\frac 12}\\=(2-e^2)^{\frac 12}(1+1-e^2)\\=(2-e^2)^{\frac 12}\times (2-e^2)\\=(2-e^2)^{\frac 32}\,\,\,\text{(proved)}$

$\,\,9.\,\,$ If $\,\,a^2\sec^2\alpha-b^2\tan^2\alpha=c^2,\,\,$ show that, $\,\,\sin\alpha=\pm\sqrt{\frac{c^2-a^2}{c^2-b^2}}.$

Sol. $\,\,a^2\sec^2\alpha-b^2\tan^2\alpha=c^2\\ \Rightarrow \frac{a^2}{\cos^2\alpha}-b^2\frac{\sin^2\alpha}{\cos^2\alpha}=c^2 \\ \Rightarrow \frac{a^2-b^2\sin^2\alpha}{\cos^2\alpha}=c^2 \\ \Rightarrow a^2-b^2\sin^2\alpha=c^2(1-\sin^2\alpha) \\ \Rightarrow (c^2-b^2)\sin^2\alpha=c^2-a^2 \\ \Rightarrow \sin^2\alpha=\frac{c^2-a^2}{c^2-b^2} \\ \Rightarrow \sin\alpha=\pm\sqrt{\frac{c^2-a^2}{c^2-b^2}}\,\,\text{(proved)}$ 

Click here  to download various PDFs of S.N. Dey Maths Solution.