Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If root 1-x^2 + 1-y^2 = a(x-y) prove that dy/dx = root 1-y^2

If \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y) prove that \frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}

Answers (1)

Solution

Given: \sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)

\begin{array}{l} \text { Put } x=\sin \alpha \text { and } y=\sin \beta \ldots(i) \\ \sqrt{1-(\sin \alpha)^{2}}+\sqrt{1-(\sin \beta)^{2}}=a(\sin \alpha-\sin \beta) \end{array}
\Rightarrow \sqrt{1-\sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \beta}=a(\sin \alpha-\sin \beta)
Now, we know that \sin ^{2} \theta+\cos ^{2} \theta=1
\\\Rightarrow \sqrt{\cos ^{2} \alpha}+\sqrt{\cos ^{2} \beta}=a(\sin \alpha-\sin \beta) \\\Rightarrow \cos \alpha+\cos \beta=a(\sin \alpha-\sin \beta) \ldots(i i)
Now, we use some trigonometry formulas,
\\ \cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)

\\ \sin \alpha-\sin \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)

\begin{aligned} &\text { So, eq.(ii) become }\\ &2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=2 \mathrm{a} \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\\ &\Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)=\mathrm{a} \sin \left(\frac{\alpha-\beta}{2}\right)\\ &\Rightarrow \mathrm{a}=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\sin \left(\frac{\alpha-\beta}{2}\right)}\\ &\Rightarrow \mathrm{a}=\cot \left(\frac{\alpha-\beta}{2}\right) \end{aligned}

\\\Rightarrow \cot ^{-1} \mathrm{a}=\frac{\alpha-\beta}{2} \\\Rightarrow 2 \cot ^{-1} a=\alpha-\beta....(i i) \\

Now, from eq.(i), we have 

\\\alpha=\sin ^{-1} x and \beta=\sin ^{-1} y

Now, put the value of \alpha and \beta in eq. (iii), we get 

\\2 \cot ^{-1} a=\sin ^{-1} x-\sin ^{-1} y \\or \sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a

On differentiating above with respect to x, we get 

\\\frac{d}{d x}\left(\sin ^{-1} x-\sin ^{-1} y\right)=\frac{d}{d x}\left(2 \cot ^{-1} a\right) \\\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{y}=2 \frac{\mathrm{d}}{\mathrm{dx}} \cot ^{-1} \mathrm{a}

\\ \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0\left[\because \frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right] \\ \Rightarrow \frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}

Hence Proved

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question