RD Chapter 4 Inverse Trigonometric Functions Ex 4.4 Solutions
Question - 1 : - Find the principal value of each of the following:
(i) sec-1 (-√2)
(ii) sec-1 (2)
(iii) sec-1 (2 sin (3π/4))
(iv) sec-1 (2 tan (3π/4))
Answer - 1 : -
(i) Given sec-1 (-√2)
Now let y = sec-1 (-√2)
Sec y = -√2
We know that sec π/4 =√2
Therefore, -sec (π/4)= -√2
= sec (π – π/4)
= sec (3π/4)
Thus the range ofprincipal value of sec-1 is [0, π] – {π/2}
And sec (3π/4) = – √2
Hence the principalvalue of sec-1 (-√2) is 3π/4
(ii) Given sec-1 (2)
Let y = sec-1 (2)
Sec y = 2
= Sec π/3
Therefore the range ofprincipal value of sec-1 is [0, π] – {π/2} and sec π/3 = 2
Thus the principalvalue of sec-1 (2) is π/3
(iii) Given sec-1 (2sin (3π/4))
But we know that sin(3π/4) = 1/√2
Therefore 2 sin (3π/4)= 2 × 1/√2
2 sin (3π/4) = √2
Therefore bysubstituting above values in sec-1 (2 sin (3π/4)), we get
Sec-1 (√2)
Let Sec-1 (√2)= y
Sec y = √2
Sec (π/4) = √2
Therefore range ofprincipal value of sec-1 is [0, π] – {π/2} and sec (π/4) = √2
Thus the principalvalue of sec-1 (2 sin (3π/4)) is π/4.
(iv) Given sec-1 (2tan (3π/4))
But we know that tan(3π/4) = -1
Therefore, 2 tan(3π/4) = 2 × -1
2 tan (3π/4) = -2
By substituting thesevalues in sec-1 (2 tan (3π/4)), we get
Sec-1 (-2)
Now let y = Sec-1 (-2)
Sec y = – 2
– sec (π/3) = -2
= sec (π – π/3)
= sec (2π/3)
Therefore the range ofprincipal value of sec-1 is [0, π] – {π/2} and sec (2π/3) = -2
Thus, the principalvalue of sec-1 (2 tan (3π/4)) is (2π/3).
Question - 2 : - Find the principalvalue of each of the following:
Answer - 2 : -
Question - 3 : - Find the domain of
sec-1 (3x - 1)
Answer - 3 : -
Question - 4 : - Find the domain of
sec-1 x - tan-1x
Answer - 4 : -