Question

If x = cos A+ sin A and y= cos A- sin A, then √x/y + √y/x is​

Answer 1

xcos(a+y)=cosy,

Which implies x=

cos(a+y)

cosy

Now differentiate on both sides , we get

dx=

cos

2

(a+y)

cos(a+y)(−siny)−cosy(−sin(a+y))

dy=

cos

2

(a+y)

sin(a+y−y)

dy=

cos

2

(a+y)

sina

dy

⇒

dx

dy

=

sina

cos

2

(a+y)

⇒

dx

2

d

2

y

=

sina

2cos(a+y)(−sin(a+y)

×

dx

dy

=

sina

−sin2(a+y)

×

dx

dy

Therefore, sina

dx

2

d

2

y

+sin2(a+y)

dx

dy

=0

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Answer 2

Given,

$\[\begin{array}{l}x = \cos A + \sin A\\y = \cos A - \sin A\end{array}\]$

Solution,

$\[ = \sqrt {\frac{x}{y}} + \sqrt {\frac{y}{x}} \]$

$\[ = \frac{{x + y}}{{\sqrt x \sqrt y }}\]$

$\[ = \frac{{\left( {\cos A + \sin A} \right) + \left( {\cos A - \sin A} \right)}}{{\sqrt {\left( {\cos A + \sin A} \right)\left( {\cos A - \sin A} \right)} }}\]$

$\[ = \frac{{2\cos A}}{{\sqrt {\left( {{{\cos }^2}A - {{\sin }^2}A} \right)} }}\]$

$\[ = \frac{{2\cos A}}{{\sqrt {\cos 2A} }}\]$

$\[ = 2\cos A\sqrt {\sec 2A} \]$

Hence the value is $\[2\cos A\sqrt {\sec 2A} \]$