Evaluate $\sin^{-1}\left(\sin\dfrac{3\pi}{4}\right)+\cos\left(\cos^{-1}\pi\right)+\tan^{-1}1$.

$\dfrac{3\pi}{2}$ (as intended by the source; see note on $\cos^{-1}\pi$)

CBSE Class 12 · 2023 · Inverse Trigonometric FunctionsShort Answer

Evaluate

sin^{- 1} (sin \frac{3 π}{4}) + cos (cos^{- 1} π) + tan^{- 1} 1

Answer: $\frac{3 π}{2}$ (as intended by the source; see note on $cos^{- 1} π$ )

sin^{- 1} (sin \frac{3 π}{4})

: since

\frac{3 π}{4} \in / [- \frac{π}{2}, \frac{π}{2}]

, use

sin \frac{3 π}{4} = sin (π - \frac{3 π}{4}) = sin \frac{π}{4}

, giving

\frac{π}{4}

tan^{- 1} 1 = \frac{π}{4}

sin^{- 1} (sin \frac{3 π}{4}) + tan^{- 1} 1 = \frac{π}{4} + \frac{π}{4} = \frac{π}{2}

Note (source defect):

cos^{- 1} π

is \emph{undefined}, because the domain of

cos^{- 1}

[- 1, 1]

and

π > 1

. The term

cos (cos^{- 1} π)

therefore does not exist, so the question as printed is not well-posed. Only the formal identity

cos (cos^{- 1} t) = t

(valid for

t \in [- 1, 1]

) applied improperly to

t = π

gives

π

; with that reading,

\frac{π}{4} + π + \frac{π}{4} = \frac{3 π}{2}

\frac{3 π}{2} (with cos^{- 1} π undefined)

CBSE Class 12 Board · 2023 · Subjective