Question

Cot-1x - cot-1y = 0 and cos-1x +cos-1y = pi/2 . x+y=

Answer 1

By adding both equations,

we get

cot-1x = pi/4

x = 1

putting in any of above equations we get

y = 1

Thus x + y = 2

Answer 2

The value of $x + y$ is $\sqrt{2}$.

Step-by-step explanation:

Given,

$cot^{-1}x - cot^{-1}y = 0$

$cot^{-1}x + cot^{-1}y = \frac{\pi }{2}$

We have to find $x + y$.

∴ $cot^{-1}x - cot^{-1}y = 0$

⇒ $cot^{-1}x = cot^{-1}y$

⇒ $x = y$

again,

∴ $cot^{-1}x + cot^{-1}y = \frac{\pi }{2}$

⇒ $cot^{-1}x + cot^{-1}x = \frac{\pi }{2}$

⇒  $2cot^{-1}x = \frac{\pi }{2}$

⇒  $cot^{-1}x = \frac{\pi }{4}$

⇒ $x = \frac{1}{\sqrt{2} }$

∴ x = y  we get

⇒ $y = \frac{1}{\sqrt{2} }$

Now, according to the problem

$x + y = \frac{1}{\sqrt{2} } + \frac{1}{\sqrt{2} }$

⇒ $x + y = \frac{2}{\sqrt{2} }$

⇒  $x + y = \frac{2 \times \sqrt{2}}{\sqrt{2}\times \sqrt{2} }$

⇒ $x + y = \frac{2 \sqrt{2}}{2}$

⇒ $x + y = \sqrt{2}$

Hence, the value of $x + y$ is $\sqrt{2}$.

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