RD Chapter 4 Inverse Trigonometric Functions Ex 4.5 Solutions
Question - 1 : - Find the principal values of each of the following:
(i) cosec-1 (-√2)
(ii) cosec-1 (-2)
(iii) cosec-1 (2/√3)
(iv) cosec-1 (2 cos (2π/3))
Answer - 1 : -
(i) Given cosec-1 (-√2)
Let y = cosec-1 (-√2)
Cosec y = -√2
– Cosec y = √2
– Cosec (π/4) = √2
– Cosec (π/4) = cosec(-π/4) [since –cosec θ = cosec (-θ)]
The range of principalvalue of cosec-1 [-π/2, π/2] – {0} and cosec (-π/4) = – √2
Cosec (-π/4) = – √2
Therefore theprincipal value of cosec-1 (-√2) is – π/4
(ii) Given cosec-1 (-2)
Let y = cosec-1 (-2)
Cosec y = -2
– Cosec y = 2
– Cosec (π/6) = 2
– Cosec (π/6) = cosec(-π/6) [since –cosec θ = cosec (-θ)]
The range of principalvalue of cosec-1 [-π/2, π/2] – {0} and cosec (-π/6) = – 2
Cosec (-π/6) = – 2
Therefore theprincipal value of cosec-1 (-2) is – π/6
(iii) Given cosec-1 (2/√3)
Let y = cosec-1 (2/√3)
Cosec y = (2/√3)
Cosec (π/3) = (2/√3)
Therefore range ofprincipal value of cosec-1 is [-π/2, π/2] – {0} and cosec (π/3)= (2/√3)
Thus, the principalvalue of cosec-1 (2/√3) is π/3
(iv) Given cosec-1 (2cos (2π/3))
But we know that cos(2π/3) = – ½
Therefore 2 cos (2π/3)= 2 × – ½
2 cos (2π/3) = -1
By substituting thesevalues in cosec-1 (2 cos (2π/3)) we get,
Cosec-1 (-1)
Let y = cosec-1 (-1)
– Cosec y = 1
– Cosec (π/2) = cosec(-π/2) [since –cosec θ = cosec (-θ)]
The range of principalvalue of cosec-1 [-π/2, π/2] – {0} and cosec (-π/2) = – 1
Cosec (-π/2) = – 1
Therefore theprincipal value of cosec-1 (2 cos (2π/3)) is – π/2
Question - 2 : - Find the set of valuesof
Answer - 2 : -
Question - 3 : - For the principalvalue evaluate the following:
Answer - 3 : - 1.
2.
3.
4.