Question -
Answer -
(i) Given sec-1┬а(-тИЪ2)
Now let y = sec-1┬а(-тИЪ2)
Sec y = -тИЪ2
We know that sec ╧А/4 =тИЪ2
Therefore, -sec (╧А/4)= -тИЪ2
= sec (╧А тАУ ╧А/4)
= sec (3╧А/4)
Thus the range ofprincipal value of sec-1┬аis [0, ╧А] тАУ {╧А/2}
And sec (3╧А/4) = тАУ тИЪ2
Hence the principalvalue of sec-1┬а(-тИЪ2) is 3╧А/4
(ii) Given sec-1┬а(2)
Let y = sec-1┬а(2)
Sec y = 2
= Sec ╧А/3
Therefore the range ofprincipal value of sec-1┬аis [0, ╧А] тАУ {╧А/2} and sec ╧А/3 = 2
Thus the principalvalue of sec-1┬а(2) is ╧А/3
(iii) Given sec-1┬а(2sin (3╧А/4))
But we know that sin(3╧А/4) = 1/тИЪ2
Therefore 2 sin (3╧А/4)= 2 ├Ч 1/тИЪ2
2 sin (3╧А/4) = тИЪ2
Therefore bysubstituting above values in sec-1┬а(2 sin (3╧А/4)), we get
Sec-1┬а(тИЪ2)
Let Sec-1┬а(тИЪ2)= y
Sec y = тИЪ2
Sec (╧А/4) = тИЪ2
Therefore range ofprincipal value of sec-1┬аis [0, ╧А] тАУ {╧А/2} and sec (╧А/4) = тИЪ2
Thus the principalvalue of sec-1┬а(2 sin (3╧А/4)) is ╧А/4.
(iv) Given sec-1┬а(2tan (3╧А/4))
But we know that tan(3╧А/4) = -1
Therefore, 2 tan(3╧А/4) = 2 ├Ч -1
2 tan (3╧А/4) = -2
By substituting thesevalues in sec-1┬а(2 tan (3╧А/4)), we get
Sec-1┬а(-2)
Now let y = Sec-1┬а(-2)
Sec y = тАУ 2
тАУ sec (╧А/3) = -2
= sec (╧А тАУ ╧А/3)
= sec (2╧А/3)
Therefore the range ofprincipal value of sec-1┬аis [0, ╧А] тАУ {╧А/2} and sec (2╧А/3) = -2
Thus, the principalvalue of sec-1┬а(2 tan (3╧А/4)) is (2╧А/3).