The value of $\tan ^{-1}\left(\frac{\cos \left(\frac{15 \pi}{4}\right)-1}{\sin \left(\frac{\pi}{4}\right)}\right)$ is equal to

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Inverse Trigonometric Functions

The value of

tan^{- 1} (\frac{c o s ( \frac{15 π}{4} ) - 1}{s i n ( \frac{π}{4} )})

is equal to

Answer: (B)

Simplify

\frac{15 π}{4} = 4 π - \frac{π}{4}

, so

cos (\frac{15 π}{4}) = cos (- \frac{π}{4}) = cos \frac{π}{4} = \frac{1}{2}

. Also

sin \frac{π}{4} = \frac{1}{2}

. So the expression becomes

tan^{- 1} (\frac{\frac{1}{2} - 1}{\frac{1}{2}}) = tan^{- 1} (1 - 2)

. Since

1 - 2 \approx - 0.414 = - tan \frac{π}{8}

, we have

tan^{- 1} (1 - 2) = - \frac{π}{8}

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