Question

`cos[tan^-1  1/3 + tan^-1  1/2]` = ______

Options

• `1/sqrt2`

• `sqrt3/2`

• `1/2`

• `pi/4`

Answer 1

`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`