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Question: If tan<sup>–1</sup>\(\frac{x + 1}{x - 1}\) + tan<sup>–1</sup>\(\frac{x - 1}{x}\) = tan<sup>–1</sup>(...

If tan–1x+1x−1\frac{x + 1}{x - 1} + tan–1x−1x\frac{x - 1}{x} = tan–1(–7) + p then x=

A

2

B

3

C

4

D

None of these

Answer

2

Explanation

Solution

[cot–1 x] = 0, 1, 2, 3, [tan–1 x] = –2, –1, 0, 1

Case (i) [cot–1 x] = [tan–1 x] = 1

x Ī (cot 2, cot 1], & x Ī [tan 1, )

\ x Īf as cot 1 < tan 1

(ii) [cot–1 x] = 3, [tan–1 x] = – 1

x Ī (–, cot 3], x Ī [– tan1, 0]

\ x Īf as cot 3 < – tan 1

(iii) [cot–1 x] = 2, [tan–1 x] = 0

xĪ (cot 3, cot 2], x Ī[0, tan 1)

xĪf as cot 2 < 0

So no solution