1. Write the domain of f(x) = sec-x. 2. Write the domain of f(x) = cos2x. 3. Write the domain of f(x) = tan-1x. 4. Write the principal value branch of f(x) = sin 2x. 5. Write the domain of f(x) = sin-1x. 6. Write the principal value branch or range of cos-lx. 7. Find the principal value of cot--1 (-3) 8. Find the principal value of cot-1 (). 9. Find the principal value of cos-- (-3). 10. Find the principal value of cos-1 (-) 11. Find the principal value of cosec-(-v2). 12. Write the set of values of x for which 2tan-1x = tan-12% 13. Write the set of values of x for which 2tan-2x = cos-1 1-x? 14. Find the value of sin-2 (sin *). 15. Find the value of sin–2 (sin). 16. Find the value of cot (tan-1x + cot-1x). 17. Find the value of cos (sec-1x + cosec-1x).\*\ > 1. holds. 1-x21 14x2 holds.
Answers
Answer:
Step-by-step explanation:
Solution
SolutionHere, F(x)=y=sec
SolutionHere, F(x)=y=sec−1
SolutionHere, F(x)=y=sec−1 x
SolutionHere, F(x)=y=sec−1 xSo, x=secy
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is R−(−1,1)
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is R−(−1,1)
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is R−(−1,1) or we can say Domain of F(x)=
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is R−(−1,1) or we can say Domain of F(x)=sec
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is R−(−1,1) or we can say Domain of F(x)=sec −1
SolutionHere, F(x)=y=sec−1 xSo, x=secynow the Range of secy is domain of sec −1 x i.e secy is possible in R−(−1,1) So the Range of secy is R−(−1,1) .. And domain of F(x)=sec −1 x is R−(−1,1) or we can say Domain of F(x)=sec −1 is (−∞,−1] U[1,∞)
Step-by-step explanation:
Taking sec on both sides we get,
sec(x)=−2
But, the principal value of cos(2π3)
is −12
i.e., the principal value of sec(2π3) is −2
.
⇒sec(2π3)=−2
∴x=2π3
Hence, the principal value of sec−1(−2) is 2π3 .