Question -
Answer -
(i) Given cosec-1┬а(-тИЪ2)
Let y = cosec-1┬а(-тИЪ2)
Cosec y = -тИЪ2
тАУ Cosec y = тИЪ2
тАУ Cosec (╧А/4) = тИЪ2
тАУ Cosec (╧А/4) = cosec(-╧А/4) [since тАУcosec ╬╕ = cosec (-╬╕)]
The range of principalvalue of cosec-1┬а[-╧А/2, ╧А/2] тАУ {0} and cosec (-╧А/4) = тАУ тИЪ2
Cosec (-╧А/4) = тАУ тИЪ2
Therefore theprincipal value of cosec-1┬а(-тИЪ2) is тАУ ╧А/4
(ii) Given cosec-1┬а(-2)
Let y = cosec-1┬а(-2)
Cosec y = -2
тАУ Cosec y = 2
тАУ Cosec (╧А/6) = 2
тАУ Cosec (╧А/6) = cosec(-╧А/6) [since тАУcosec ╬╕ = cosec (-╬╕)]
The range of principalvalue of cosec-1┬а[-╧А/2, ╧А/2] тАУ {0} and cosec (-╧А/6) = тАУ 2
Cosec (-╧А/6) = тАУ 2
Therefore theprincipal value of cosec-1┬а(-2) is тАУ ╧А/6
(iii) Given cosec-1┬а(2/тИЪ3)
Let y = cosec-1┬а(2/тИЪ3)
Cosec y = (2/тИЪ3)
Cosec (╧А/3) = (2/тИЪ3)
Therefore range ofprincipal value of cosec-1┬аis [-╧А/2, ╧А/2] тАУ {0} and cosec (╧А/3)= (2/тИЪ3)
Thus, the principalvalue of cosec-1┬а(2/тИЪ3) is ╧А/3
(iv) Given cosec-1┬а(2cos (2╧А/3))
But we know that cos(2╧А/3) = тАУ ┬╜
Therefore 2 cos (2╧А/3)= 2 ├Ч тАУ ┬╜
2 cos (2╧А/3) = -1
By substituting thesevalues in cosec-1┬а(2 cos (2╧А/3)) we get,
Cosec-1┬а(-1)
Let y = cosec-1┬а(-1)
тАУ Cosec y = 1
тАУ Cosec (╧А/2) = cosec(-╧А/2) [since тАУcosec ╬╕ = cosec (-╬╕)]
The range of principalvalue of cosec-1┬а[-╧А/2, ╧А/2] тАУ {0} and cosec (-╧А/2) = тАУ 1
Cosec (-╧А/2) = тАУ 1
Therefore theprincipal value of cosec-1┬а(2 cos (2╧А/3)) is тАУ ╧А/2