The Total solution for NCERT class 6-12
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 - 1 : -
(i) Given cot-1(-√3)
Let y = cot-1(-√3)
– Cot (π/6) = √3
= Cot (π – π/6)
= cot (5π/6)
The range of principalvalue of cot-1 is (0, π) and cot (5 π/6) = – √3
Thus, the principalvalue of cot-1 (- √3) is 5π/6
(ii) Given Cot-1(√3)
Let y = cot-1(√3)
Cot (π/6) = √3
The range of principalvalue of cot-1 is (0, π) and
Thus, the principalvalue of cot-1 (√3) is π/6
(iii) Given 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)
The range of principalvalue of cot-1(0, π) and cot (2π/3) = – 1/√3
Therefore theprincipal value of cot-1(-1/√3) is 2π/3
(iv) Given cot-1(tan3π/4)
But we know that tan3π/4 = -1
By substituting thisvalue in cot-1(tan 3π/4) we get
Cot-1(-1)
Now, let y = cot-1(-1)
Cot y = (-1)
– Cot (π/4) = 1
= Cot (π – π/4)
= cot (3π/4)
The range of principalvalue of cot-1(0, π) and cot (3π/4) = – 1
Therefore theprincipal value of cot-1(tan 3π/4) is 3π/4
Find the domain of f(x) = cot x + cot-1 x.
Answer - 2 : -
Evaluate each of thefollowing:
Answer - 3 : -
Answer - 4 : -
Answer - 5 : -
Answer - 6 : -