Question

show solution no one direct answer
I also know the answer but
I want solution
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Answer 1

It's too simple dear

Just do this and nothing ,

a = [ (x^2 + y^2)^1/2 + (x^2 - y^2)^1/2 ] / [ (x^2 + y^2)^1/2 - (x^2 - y^2)^1/2 ]

So

a * [ (x^2 + y^2)^1/2 - (x^2 - y^2)^1/2 ] = [ (x^2 + y^2)^1/2 + (x^2 - y^2)^1/2 ]

a*(x^2 + y^2)^1/2 - a*(x^2 - y^2)^1/2 = (x^2 + y^2)^1/2 + (x^2 - y^2)^1/2

(a - 1)*(x^2 + y^2)1/2 =(a + 1) *(x^2 - y^2)^1/2

Now square both sides

(a - 1)^2 * (x^2 + y^2) = (a + 1)^2 * (x^2 - y^2)

Expand the equation

(a^2x^2) + a^2y^2 - 2ax^2 - (2ay^2) + (x^2) + y^2 = (a^2x^2) - a^2y^2 + 2ax^2 - (2ay^2) + (x^2) - y^2

Terms in brackets are canceled

a^2y^2 - 2ax^2 + y^2 = - a^2y^2 + 2ax^2 - y^2

So ,

2 a^2y^2 - 2 (2ax^2) + 2 y^2 = 0

a^2y^2 - (2ax^2) + y^2 = 0

a^2y^2 = (2ax^2) + y^2

a^2 = [ (2ax^2) - y^2 ] / y^2

= 2ax^2 / y^2 - 1

a^2 + 1 = 2ax^2 / y^2

( a^2 + 1)/a = 2x^2 / y^2

Now a + 1 / a = ( a^2 + 1)/a

= 2x^2 / y^2

So option (D) is correct

Answered by an IIT JEE ASPIRANT and all India mathematics OLYMPIAD TOPPER in class 10th

Answer 2

Solution :-

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

To solve first make the question into simpler one.

Let us assume

• √( x² + y² ) = p

• √(x² - y² ) = q

Substituting the assumed values in the equation

$\implies \sf a = \dfrac{p + q}{p - q}$

Now, a + 1/a

$\implies \displaystyle{ \sf a + \frac{1}{a} = \frac{p + q}{p - q} + \frac{1}{ \frac{p + q}{p - q} } }$

$\implies \sf a + \dfrac{1}{a} = \dfrac{p + q}{p - q} + \dfrac{p - q}{p + q}$

Taking LCM

$\implies \sf a + \dfrac{1}{a} = \dfrac{p + q(p + q) + p - q(p - q)}{(p - q)(p + q)}$

$\implies \sf a + \dfrac{1}{a} = \dfrac{(p + q)^{2} + (p - q)^{2} }{ {p}^{2} - {q}^{2} }$

[ Because, (p - q)(p + q) = p² - q² ]

$\implies \sf a + \dfrac{1}{a} = \dfrac{ {p}^{2} + {q}^{2} + 2pq + {p}^{2} + {q}^{2} - 2pq }{ {p}^{2} - {q}^{2} }$

[ Because (p + q)² = p² + q² + 2pq , (p - q)² = p² + q² - 2pq ]

$\implies \sf a + \dfrac{1}{a} = \dfrac{ 2{p}^{2} + 2{q}^{2} }{ {p}^{2} - {q}^{2} }$

Now substituting p = √(x² + y²) and q = √(x² - y²)

$\implies \sf a + \dfrac{1}{a} = \dfrac{ 2 \big({ \sqrt{ {x}^{2} + {y}^{2} } } \big)^{2} + 2 \big( \sqrt{ {x}^{2} - {y}^{2} } \big) ^{2} }{ \big({ \sqrt{ {x}^{2} + {y}^{2} } } \big)^{2} - \big({ \sqrt{ {x}^{2} - {y}^{2} } } \big)^{2} }$

$\implies \sf a + \dfrac{1}{a} = \dfrac{ 2 ( {x}^{2} + {y}^{2} ) + 2 ( {x}^{2} - {y}^{2}) }{ {x}^{2} + {y}^{2} - ({x}^{2} - {y}^{2}) }$

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

$\implies \sf a + \dfrac{1}{a} = \dfrac{ 4{x}^{2} }{ 2 {y}^{2} }$

$\implies \sf a + \dfrac{1}{a} = \dfrac{ 2{x}^{2} }{ {y}^{2} }$

Hence, the value of a + 1/a is 2x²/y²