Cartesian plane

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The area of the triangle formed by the lines $y = x, y = 0 and x = \sin^{- 1} (a^{4} + 1) + \cos^{- 1} (a^{4} + 1) - \tan^{- 1} (a^{4} + 1)$ is sq. units, where $k$ =

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Solution

We observe that $a^{4} + 1 \geq 1$ for all $a \in R$ Therefore $x = \sin^{- 1} (a^{4} + 1) + \cos^{- 1} (a^{4} + 1) - \tan^{- 1} (a^{4} + 1)$ is meaning it for $a = 0$ and in that case it becomes $y = \frac{π}{4}$ . Thus, the given lines $y = x, y = 0 and x = \frac{π}{4}$ These lines form a right triangle area $\frac{π^{2}}{32}$ sq. umts

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Cartesian plane

Remember concepts with our Masterclasses.

The area of the triangle formed by the lines y=x, y=0 and x=sin−1a4+1+cos−1⁡a4+1−tan−1⁡a4+1 is sq. units, where k =

We observe that a4+1≥1 for all a∈R Therefore x=sin−1⁡a4+1+cos−1⁡a4+1−tan−1⁡a4+1 is meaning it for a= 0 and in that case it becomes y=π4. Thus, the given lines y=x,y=0 and x=π4 These lines form a right triangle area π232 sq. umts

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free study material

The area of the triangle formed by the lines $y = x, y = 0 and x = \sin^{- 1} (a^{4} + 1) + \cos^{- 1} (a^{4} + 1) - \tan^{- 1} (a^{4} + 1)$ is sq. units, where $k$ =