If cos^{-1}left(frac{2}{3x}right) + cos^{-1}l | vedclass

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(A) Given $\cos^{-1}\left(\frac{2}{3x}\right) + \cos^{-1}\left(\frac{3}{4x}\right) = \frac{\pi}{2}$.We know that $\cos^{-1}(A) + \cos^{-1}(B) = \cos^{

Vedclass · Accepted Answer

$\frac{\sqrt{145}}{12}$

Vedclass · Answer

(A) Given $\cos^{-1}\left(\frac{2}{3x}ight) + \cos^{-1}\left(\frac{3}{4x}ight) = \frac{\pi}{2}$.We know that $\cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}\left(AB - \sqrt{1-A^2}\sqrt{1-B^2}ight)$.So,$\cos^{-1}\left(\frac{2}{3x} \cdot \frac{3}{4x} - \sqrt{1-\frac{4}{9x^2}}\sqrt{1-\frac{9}{16x^2}}ight) = \frac{\pi}{2}$.Taking $\cos$ on both sides,$\frac{6}{12x^2} - \sqrt{\frac{9x^2-4}{9x^2}}\sqrt{\frac{16x^2-9}{16x^2}} = \cos\left(\frac{\pi}{2}ight) = 0$.$\frac{1}{2x^2} = \frac{\sqrt{9x^2-4}\sqrt{16x^2-9}}{12x^2}$.Multiplying by $12x^2$,we get $6 = \sqrt{9x^2-4}\sqrt{16x^2-9}$.Squaring both sides,$36 = (9x^2-4)(16x^2-9) = 144x^4 - 81x^2 - 64x^2 + 36$.$144x^4 - 145x^2 = 0$.$x^2(144x^2 - 145) = 0$.Since $x > \frac{3}{4}$,$x^2 
eq 0$,so $144x^2 = 145$,which gives $x = \frac{\sqrt{145}}{12}$.

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

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