We are given
sin − 1 ( x 5 ) + csc − 1 ( 5 4 ) = π 2 . \sin^{-1}\left(\frac{x}{5}\right)+\csc^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}. sin − 1 ( 5 x ) + csc − 1 ( 4 5 ) = 2 π . We need to find x x x
First, evaluate csc − 1 ( 5 4 ) \csc^{-1}\left(\frac{5}{4}\right) csc − 1 ( 4 5 )
Let
θ = csc − 1 ( 5 4 ) . \theta=\csc^{-1}\left(\frac{5}{4}\right). θ = csc − 1 ( 4 5 ) . Then
csc θ = 5 4 ⇒ sin θ = 4 5 . \csc \theta=\frac{5}{4}
\quad\Rightarrow\quad
\sin\theta=\frac{4}{5}. csc θ = 4 5 ⇒ sin θ = 5 4 . Since the principal value of csc − 1 y \csc^{-1}y csc − 1 y [ − π 2 , 0 ) ∪ ( 0 , π 2 ] \left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right] [ − 2 π , 0 ) ∪ ( 0 , 2 π ] 5 4 > 1 \frac{5}{4}>1 4 5 > 1
θ = sin − 1 ( 4 5 ) . \theta=\sin^{-1}\left(\frac{4}{5}\right). θ = sin − 1 ( 5 4 ) . So,
csc − 1 ( 5 4 ) = sin − 1 ( 4 5 ) . \csc^{-1}\left(\frac{5}{4}\right)=\sin^{-1}\left(\frac{4}{5}\right). csc − 1 ( 4 5 ) = sin − 1 ( 5 4 ) .
Substitute this into the given equation:
sin − 1 ( x 5 ) + sin − 1 ( 4 5 ) = π 2 . \sin^{-1}\left(\frac{x}{5}\right)+\sin^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}. sin − 1 ( 5 x ) + sin − 1 ( 5 4 ) = 2 π .
Rearranging,
sin − 1 ( x 5 ) = π 2 − sin − 1 ( 4 5 ) . \sin^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin^{-1}\left(\frac{4}{5}\right). sin − 1 ( 5 x ) = 2 π − sin − 1 ( 5 4 ) . Now use the identity
π 2 − sin − 1 ( t ) = cos − 1 ( t ) . \frac{\pi}{2}-\sin^{-1}(t)=\cos^{-1}(t). 2 π − sin − 1 ( t ) = cos − 1 ( t ) . Thus,
sin − 1 ( x 5 ) = cos − 1 ( 4 5 ) . \sin^{-1}\left(\frac{x}{5}\right)=\cos^{-1}\left(\frac{4}{5}\right). sin − 1 ( 5 x ) = cos − 1 ( 5 4 ) . But if sin θ = 4 5 \sin\theta=\frac{4}{5} sin θ = 5 4
cos θ = 3 5 . \cos\theta=\frac{3}{5}. cos θ = 5 3 . Hence,
cos − 1 ( 4 5 ) is not directly needed; instead take sine on both sides of sin − 1 ( x 5 ) = π 2 − sin − 1 ( 4 5 ) . \cos^{-1}\left(\frac{4}{5}\right)\text{ is not directly needed; instead take sine on both sides of }
\sin^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin^{-1}\left(\frac{4}{5}\right). cos − 1 ( 5 4 ) is not directly needed; instead take sine on both sides of sin − 1 ( 5 x ) = 2 π − sin − 1 ( 5 4 ) . So,
x 5 = sin ( π 2 − sin − 1 ( 4 5 ) ) = cos ( sin − 1 ( 4 5 ) ) . \frac{x}{5}=
\sin\left(\frac{\pi}{2}-\sin^{-1}\left(\frac{4}{5}\right)\right)
=\cos\left(\sin^{-1}\left(\frac{4}{5}\right)\right). 5 x = sin ( 2 π − sin − 1 ( 5 4 ) ) = cos ( sin − 1 ( 5 4 ) ) .
If
sin α = 4 5 , \sin\alpha=\frac{4}{5}, sin α = 5 4 , then
cos α = 1 − sin 2 α = 1 − 16 25 = 9 25 = 3 5 . \cos\alpha=\sqrt{1-\sin^2\alpha}
=\sqrt{1-\frac{16}{25}}
=\sqrt{\frac{9}{25}}
=\frac{3}{5}. cos α = 1 − sin 2 α = 1 − 25 16 = 25 9 = 5 3 . Therefore,
x 5 = 3 5 . \frac{x}{5}=\frac{3}{5}. 5 x = 5 3 . So,
x = 3. x=3. x = 3.
Checking options:
A: 4 4 4
B: 5 5 5
C: 1 1 1
D: 3 3 3
Therefore, the correct answer is
3 . \boxed{3}. 3 .