If $\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2}, x>\frac{3}{4}$ then $x$ is equal to:

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Inverse Trigonometric Functions

cos^{- 1} (\frac{2}{3 x}) + cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2}, x > \frac{3}{4}

then

x

is equal to:

Answer: (A)

Use

cos^{- 1} A + cos^{- 1} B = \frac{π}{2} \Rightarrow A B = (1 - A^{2}) (1 - B^{2})

. Here

\frac{2}{3 x} \cdot \frac{3}{4 x} = \frac{1}{2 x ^{2}}

, and the right side is

\frac{( 9 x ^{2} - 4 ) ( 16 x ^{2} - 9 )}{12 x ^{2}}

. Equating gives

6 = (9 x^{2} - 4) (16 x^{2} - 9)

; squaring and simplifying,

144 x^{4} = 145 x^{2}

, so

x^{2} = \frac{145}{144}

. Since

x > \frac{3}{4}

x = \frac{145}{12}

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