The value of tan \(\frac{\pi }{8}\) is equal to1. √2 - 12. \(\frac

1.	The value of tan \(\frac{\pi }{8}\) is equal to1. √2 - 12. \(\frac{1}{{\sqrt 2 - 1}}\)3. -√2 - 14. \(\frac{1}{2}\)
Answer» Correct Answer - Option 1 : √2 - 1 Concept: \(\rm \tan (x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\) Calculation: Let x = tan \(\pi\over8\) x = tan (\({\pi\over4} - {\pi\over8}\)) x = \(\tan {\pi\over4} - \tan {\pi\over8}\over1+\tan {\pi\over4}\times\tan {\pi\over8}\) x = \(\rm 1 - x\over 1+ x\) x(1 + x) = 1 - x x² + 2x - 1 = 0 \(\rm x = {-2 \pm \sqrt{2^2-4(-1)} \over 2}\) \(\rm x = {-2 \pm 2\sqrt2 \over 2}\) x = -1 + √2 or -1 - √2 Negative value cannot be possible in first quadrant ∴ tan \(\pi\over8\) = x = √2 - 1

The value of tan \(\frac{\pi }{8}\) is equal to1. √2 - 12. \(\frac{1}{{\sqrt 2 - 1}}\)3. -√2 - 14. \(\frac{1}{2}\)

Answer» Correct Answer - Option 1 : √2 - 1

Concept:

\(\rm \tan (x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\)

Calculation:

Let x = tan \(\pi\over8\)

x = tan (\({\pi\over4} - {\pi\over8}\))

x = \(\tan {\pi\over4} - \tan {\pi\over8}\over1+\tan {\pi\over4}\times\tan {\pi\over8}\)

x = \(\rm 1 - x\over 1+ x\)

x(1 + x) = 1 - x

x² + 2x - 1 = 0

\(\rm x = {-2 \pm \sqrt{2^2-4(-1)} \over 2}\)

\(\rm x = {-2 \pm 2\sqrt2 \over 2}\)

x = -1 + √2 or -1 - √2

Negative value cannot be possible in first quadrant

∴ tan \(\pi\over8\) = x = √2 - 1