18. साजित बरोट $$ an ^{-1}\left(\frac{1}{8} ight)+ an ^{-1}\left(\frac{1}{5} ight)+ an ^{-1}\left(\frac{1}{2} ight)=\frac{\pi}{4} $$

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18. साजित बरोट $$ 	an ^{-1}\left(\frac{1}{8}ight)+	an ^{-1}\left(\frac{1}{5}ight)+	an ^{-1}\left(\frac{1}{2}ight)=\frac{\pi}{4} $$

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The given equation is:

$$ \tan^{-1}\left(\frac{1}{8}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} $$

To solve this equation, we can use the formula for the sum of inverse tangents:

$$ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $$

Let's apply this formula to the given equation.
Calculate the value by following steps:
- step0: Calculate:
  $$\arctan\left(\frac{1}{8}\right)+\arctan\left(\frac{1}{5}\right)+\arctan\left(\frac{1}{2}\right)$$
- step1: Calculate the trigonometric value:
  $$0.785398$$
The sum of the inverse tangents is approximately 0.785398.

Now, we need to check if this value is equal to $$ \frac{\pi}{4} $$. Let's calculate $$ \frac{\pi}{4} $$ and compare the two values.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\pi }{4}$$
The value of $$ \frac{\pi}{4} $$ is approximately 0.785398, which matches the sum of the inverse tangents we calculated earlier.

Therefore, the equation $$ \tan^{-1}\left(\frac{1}{8}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} $$ is true.
The equation $$ \tan^{-1}\left(\frac{1}{8}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} $$ is true.

18. साजित बरोट \( \tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{4} \)

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