Find the principal values of each of the following:(i) cot-1(-√3)(ii

1.	Find the principal values of each of the following:(i) cot-1(-√3)(ii) cot-1(√3)(iii) cot-1(-1/√3)(iv) cot-1(tan 3π/4)
Answer» (i) Given as cot^-1(-√3) Let y = cot^-1(-√3) – cot (π/6) = √3 = cot (π – π/6) = cot (5π/6) So, the range of principal value of cot^-1 is (0, π) and cot (5 π/6) = – √3 Hence, the principal value of cot^-1 (-√3) is 5π/6 (ii) Given as cot^-1(√3) Let y = cot^-1(√3) cot (π/6) = √3 So, the range of principal value of cot^-1 is (0, π) and Hence, the principal value of cot^-1 (√3) is π/6 (iii) Given as cot^-1(-1/√3) Let y = cot^-1(-1/√3) cot y = (-1/√3) – cot (π/3) = 1/√3 = cot (π – π/3) = cot (2π/3) So, the range of principal value of cot^-1(0, π) and cot (2π/3) = – 1/√3 So, the principal value of cot^-1(-1/√3) is 2π/3 (iv) Given as cot^-1(tan 3π/4) As we know that tan 3π/4 = -1 Substitute these value in cot^-1(tan 3π/4) we get cot^-1(-1) Let y = cot^-1(-1) cot y = (-1) – cot (π/4) = 1 = cot (π – π/4) = cot (3π/4) So, the range of principal value of cot^-1(0, π) and cot (3π/4) = – 1 So, the principal value of cot^-1(tan 3π/4) is 3π/4

Find the principal values of each of the following:(i) cot-1(-√3)(ii) cot-1(√3)(iii) cot-1(-1/√3)(iv) cot-1(tan 3π/4)

Answer»

(i) Given as cot^-1(-√3)

Let y = cot^-1(-√3)

– cot (π/6) = √3

= cot (π – π/6)

= cot (5π/6)

So, the range of principal value of cot^-1 is (0, π) and cot (5 π/6) = – √3

Hence, the principal value of cot^-1 (-√3) is 5π/6

(ii) Given as cot^-1(√3)

Let y = cot^-1(√3)

cot (π/6) = √3

So, the range of principal value of cot^-1 is (0, π) and

Hence, the principal value of cot^-1 (√3) is π/6

(iii) Given as cot^-1(-1/√3)

Let y = cot^-1(-1/√3)

cot y = (-1/√3)

– cot (π/3) = 1/√3

= cot (π – π/3)

= cot (2π/3)

So, the range of principal value of cot^-1(0, π) and cot (2π/3) = – 1/√3

So, the principal value of cot^-1(-1/√3) is 2π/3

(iv) Given as cot^-1(tan 3π/4)

As we know that tan 3π/4 = -1

Substitute these value in cot^-1(tan 3π/4) we get

cot^-1(-1)

Let y = cot^-1(-1)

cot y = (-1)

– cot (π/4) = 1

= cot (π – π/4)

= cot (3π/4)

So, the range of principal value of cot^-1(0, π) and cot (3π/4) = – 1

So, the principal value of cot^-1(tan 3π/4) is 3π/4