Given 0 ≤ x ≤ 2 1 then the value of tan [ sin − 1 { 2 x + 2 1 − x 2 } − sin − 1 x ] is

Put x = sin θ ⇒ tan [ sin − 1 ( sin ( θ + 4 π ) ) − θ ] = 1

Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionsProperties of ITFMedium2 minQB

MathematicsMediumsingle choice

Given $0 \leq x \leq \frac{1}{2}$ then the value of $tan [sin^{- 1} {\frac{x}{2} + \frac{1 - x ^{2}}{2}} - sin^{- 1} x]$ is

Answer:

Solution:

Put $x = sin θ \Rightarrow tan [sin^{- 1} (sin (θ + \frac{π}{4})) - θ] = 1$

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:Properties of ITF

⏱ 2mℹ️ Source: QB

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