Question -
Answer -
(i) Given cot-1(-тИЪ3)
Let y = cot-1(-тИЪ3)
тАУ Cot (╧А/6) = тИЪ3
= Cot (╧А тАУ ╧А/6)
= cot (5╧А/6)
The range of principalvalue of cot-1┬аis (0, ╧А) and cot (5 ╧А/6) = тАУ тИЪ3
Thus, the principalvalue of cot-1┬а(- тИЪ3) is 5╧А/6
(ii) Given Cot-1(тИЪ3)
Let y = cot-1(тИЪ3)
Cot (╧А/6) = тИЪ3
The range of principalvalue of cot-1┬аis (0, ╧А) and
Thus, the principalvalue of cot-1┬а(тИЪ3) is ╧А/6
(iii) Given cot-1(-1/тИЪ3)
Let y = cot-1(-1/тИЪ3)
Cot y = (-1/тИЪ3)
тАУ Cot (╧А/3) = 1/тИЪ3
= Cot (╧А тАУ ╧А/3)
= cot (2╧А/3)
The range of principalvalue of cot-1(0, ╧А) and cot (2╧А/3) = тАУ 1/тИЪ3
Therefore theprincipal value of cot-1(-1/тИЪ3) is 2╧А/3
(iv) Given cot-1(tan3╧А/4)
But we know that tan3╧А/4 = -1
By substituting thisvalue in cot-1(tan 3╧А/4) we get
Cot-1(-1)
Now, let y = cot-1(-1)
Cot y = (-1)
тАУ Cot (╧А/4) = 1
= Cot (╧А тАУ ╧А/4)
= cot (3╧А/4)
The range of principalvalue of cot-1(0, ╧А) and cot (3╧А/4) = тАУ 1
Therefore theprincipal value of cot-1(tan 3╧А/4) is 3╧А/4