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Question
`cos[tan^-1 1/3 + tan^-1 1/2]` = ______
Options
`1/sqrt2`
`sqrt3/2`
`1/2`
`pi/4`
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Solution
`cos[tan^-1 1/3 + tan^-1 1/2] = \underlinebb((1/sqrt2))`
Explanation:
Let consider, y = `cos[tan^-1 1/3 + tan^-1 1/2]`
∵ `tan^-1 a + tan^-1 b = tan^-1 ((a + b)/(1 - ab))`, if ab < 1
∴ a = `1/3`; b = `1/2`; ab = `1/6 < 1`;
y = `cos [tan^-1 ((1/3 + 1/2)/(1 - 1/3 xx 1/2))]`
Thus, y = `cos [tan^-1 (((3 + 2)/6)/((6 - 1)/6))]`
y = `cos[tan^-1 5/5]`
y = cos[tan−1(1)]
y = `cos(pi/4)`
y = cos(45°)
∴ y = `1/sqrt2`
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