sec^-1>1< (not eleven)
I have missed so much school and don't really know what to do. If you could help me out I would really appreciate it
a) Just use a scientific calculator, find the tan button and locate the 2nd function button on the calculator, press the 2nd function button then the tan button then open parentheses and type (sqrt3/3) then hit equal. Also note that sqrt3/3 = 1/sqrt3 (check with calculator for proof, or multiply (1/sqrt3) by (sqrt3/sqrt3) ), which is the ratio of tan 30 degrees in this special 30–60–90 degrees triangle and based on the image provided, the angle 30 satisfies tan^-1 (1/sqrt3).
b) I am not sure I understand the question, can you verify it please.
c)csc (x) is equal to sin (1/x) . Therefor, csc^-1 (sqrt2) = sin^-1 (1/sqrt2) , same principal as in part a): Use a calculator or NOTE that sqrt2 is part of another special triangle: and based on the diagram I gave you, the angle 45 satisfies this problem since sine is opp/hyp, and the ratio of angle 45 (sin) is 1 /sqrt2 .
d) You should try this question using a calculator and the special triangles I provided. Hint cot (x) = tan (1/x)
If you have more questions please ask!