Sum

Prove that cot^{−1}(7) + 2 cot^{−1}(3) = `pi/4`

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#### Solution

L.H.S. = cot^{−1}(7) + 2 cot^{−1}(3)

= cot^{–1}(7) + cot^{–1}(3) + cot^{–1}(3)

= `pi/2 -tan^-1(7) + pi/2 - tan^-1(3) + pi/2 - tan^-1(3)` .......`[∵ tan^(−1)x + cot^(−1)x = pi/2]`

= `(3pi)/2 - [pi + tan^-1 ((7 + 3)/(1 - 7 xx 3)) + tan^-1(3)]` .......`[∵ tan^(-1)x + tan^(-1)y = pi + tan^(-1) (x + y)/(1 - xy), "if" x, y > 0 and xy > 1]`

= `(3pi)/2 - pi - [tan^-1 (10/-20) + tan^-1(3)]`

= `pi/2 - [tan^-1 (1/2) + tan^-1(3)]`

= `pi/2 - [tan^-1(3) - tan^-1(1/2)]` .......`[∵ tan^-1(-x) = -tan^-1(x)]`

= `pi/2 - [tan^-1((3 - 1/2)/(1 + (3)(1/2)))]`

= `pi/2 - [tan^-1((5/2)/(5/2))]`

= `pi/2 - tan^-1(1)`

= `pi/2 - pi/4`

= `pi/4`

= R.H.S.

Concept: Inverse Trigonometric Functions

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