Solve the equation cos (tan-1 x) = sin (cot-1 3/4).
Given equation, cos (tan-1 x) = sin (cot-1 3/4)
Taking L.H.S,
We have, $\cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right)$
L.H.S. $=\cos \left(\tan ^{-1} x\right)$
$=\cos \left(\cos ^{-1} \frac{1}{\sqrt{x^{2}+1}}\right)$
$=\frac{1}{\sqrt{x^{2}+1}}$ $\left(\because \cos \left(\cos ^{-1} x\right)=x, x \in[-1,1]\right)$
R.H.S. $=\sin \left(\cot ^{-1} \frac{3}{4}\right)$
$=\sin \left(\sin ^{-1} \frac{4}{5}\right)$
$=\frac{4}{5}$ $\left(\because \sin \left(\sin ^{-1} x\right)=x, x \in[-1,1]\right)$
Hence, on equating the L.H.S and R.H.S,we get $\frac{1}{\sqrt{x^{2}+1}}=\frac{4}{5}$
On squaring on both sides,
16(x2 + 1) = 25
16x2 + 16 = 25
16x2 = 9
x2 = 9/16
Therefore, x = ± 3/4
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