cos[tan^-1 $\frac{1}{3}$ + tan^-1 $\frac{1}{2}$] = ______

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KollegeApply · Accepted Answer

`cos[tan^-1 $\frac{1}{3}$ + tan^-1 $\frac{1}{2}$] = \underlinebb((1/sqrt2))` Explanation: Let consider, y = cos[tan^-1 $\frac{1}{3}$ + tan^-1 $\frac{1}{2}$] ∵ `tan^-1 a + tan^-1 b = tan^-1 ((a + b)/(1 - ab)), if ab < 1 $\therefore$ a = $\frac{1}{3}$; b = $\frac{1}{2}$; ab = $\frac{1}{6}$ < 1; y = cos [tan^-1 (($\frac{1}{3}$ + $\frac{1}{2}$)/(1 - $\frac{1}{3}$ xx $\frac{1}{2}$))] Thus, y = cos [tan^-1 (((3 + 2)/6)/((6 - 1)/6))] y = cos[tan^-1 $\frac{5}{5}$] y = cos[tan−1(1)] y = cos(pi/4) y = cos($45^\circ$) $\therefore$ y = 1/sqrt2`

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