Question

$\sqrt{x {}^{2} + y {}^{2} } - y \div x - \sqrt{x {}^{2} - y {}^{2} } \div \sqrt{x {}^{2} - y {}^{2} } + x \div \sqrt{x {}^{2} + y {}^{2} } + y$
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Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\sqrt{x^2+y^2}-y\:\div \:x-\sqrt{x^2-y^2}\:\div \sqrt{x^2-y^2}+x\:\div \sqrt{x^2+y^2}+y}$

♣ ᴀɴꜱᴡᴇʀ :

$\sf{\sqrt{x^2+y^2}-\dfrac{y}{x}-\dfrac{\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}+\dfrac{x}{\sqrt{x^2+y^2}}+y=\sqrt{x^2+y^2}-\dfrac{y}{x}-1+\dfrac{x\sqrt{x^2+y^2}}{x^2+y^2}+y}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sqrt{x^2+y^2}-\dfrac{y}{x}-\dfrac{\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}+\dfrac{x}{\sqrt{x^2+y^2}}+y$

$\mathrm{Apply\:rule}\:\dfrac{a}{a}=1$

$\dfrac{\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}=1$

$=\sqrt{x^2+y^2}-\dfrac{y}{x}-1+\dfrac{x}{\sqrt{x^2+y^2}}+y$

$\dfrac{x}{\sqrt{x^{2}+y^{2}}}=\dfrac{x \sqrt{x^{2}+y^{2}}}{x^{2}+y^{2}}$

$\boxed{\sf{=\sqrt{x^2+y^2}-\dfrac{y}{x}-1+\dfrac{x\sqrt{x^2+y^2}}{x^2+y^2}+y}}$

Answer 2

thank you so much sandeep....

i will return you thanks by my second id that is anchal1217....

because this id is in pc.......

so, you can understand how difficult it is to thank......

again thanks a lot.....