{Sec^2 585°+ Sin 19π/2} ÷ {csc 840°cot(-870)-cos17π tan(-10π)}

2) -0.732050808 = -0.732

{Tan(-510°) Cot 930° Sec^2 1913 π} ÷ {(cos 7π/4 csc 9π/4

[Sin(-210)-cos(870°)]) + tan 1122π}

3) 2.976067743 = 2.98

{1/cos750° [sec420°-tan7π/6][ tan(-2π/3) + sin^2 2250]- tan^2 780°} ÷ {tan^2 675°[1/sec^2 3π/4]-sin180°}

Above you will find the three answers. Just below each you will find your problems slightly restated (a few extra parentheses type symbols) to better show how I took the desired groupings to be. The biggest principle followed was that everything to the left of a division sign was taken to be divided by everything to the right of the sign. This is ever so slightly at odds with the two-choice approach I took on your question which only had the 3) expression. But 3) still works out to be almost the same value!

And I cannot spend as much time on them as I did the other question, so I just give the answers above in decimal form rather than in “(14 + 3√3)/7” form. Sorry!

sec^2 585 = 1/(cos 585)^2 = 2

sin 19 pi / 2= sin 1710= -1

csc 840= 1/ sin 840 = 2 times the square root of 3 over 3

cot -870= 1/ tan -870= square root of 3

cos 17pi = cos 3060= -1

tan -10pi = tan 1800= 0

______________________2-1____________________

(2times the square root of 3 over 3)(square root of 3) – 1(0)

___1____ = 1 / 2 = 0.5

2-0

sorry i cannot solve the other questions because i have also other homeworks to do… after this i will now signout to yahoo..