-a2-b2-a-a2-b2-b-a2-b2-ba-a2-b2

a+√(a^2 b^2)√(a^2+b^2)+b÷√(a^2+b^2) ba √(a^2 b^2) a+√(a^2 b^2)√(a^2+b^2)+b× #160; a √(a^2 b^2)√(a^2+b^2) b a^2 (a^2 b^2)(a^2+b^2) b^2 ( ba ...

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Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Polynomial

√a2-b2+a√a2+b...

Question

$\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}$

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Solution

$\frac{a + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}$

$\frac{a + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} + b} \times \frac{a - \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - b}$

$\frac{a^{2} - (a^{2} - b^{2})}{(a^{2} + b^{2}) - b^{2}}$

${(\frac{b}{a})}^{2}$

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Similar questions

Q. If θ =

\frac{a}{b}

, then

\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}

= ?
(a)

\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}

(b)

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

(c)

\frac{a^{2}}{(a^{2} + b^{2})}

(d)

\frac{b^{2}}{(a^{2} + b^{2})}

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

The equation of the circle with center (– a, – b) and radius √a² – b² is:

Q. Simplify

\frac{a + \sqrt{a^{2} - b^{2}}}{a - \sqrt{a^{2} - b^{2}}} + \frac{a - \sqrt{a^{2} - b^{2}}}{a + \sqrt{a^{2} - b^{2}}}

Q. Show that :

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}} = \frac{b^{2}}{d^{2}}

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√a2−b2+a√a2+b2+b÷√a2+b2−ba−√a2−b2

a+√a2−b2√a2+b2+b÷√a2+b2−ba−√a2−b2a+√a2−b2√a2+b2+b×a−√a2−b2√a2+b2−ba2−(a2−b2)(a2+b2)−b2(ba)2

$\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}$