Rule: The square of an odd gives an odd & the square of an even number is always even
We must plugin values:
3(7)-4(0)=21, is CORRECT since 7•3 is 21 and -4•0 is 0
3(11)-4(3)=21 is CORRECT since 3•11=33 and -4•3=-12 so 33-12=21
3(-3)-4(3)=21 is INCORRECT since 3•-3 is -9 and -4•3 is -12, -9-12 is -21
3(-1)-4(-6)=21 is CORRECT since 3•-1 is -3 and -4•-6 is 24 and so 24-3=21
Answer:
See proof below
Step-by-step explanation
cot^2(x) - csc^2(x) = -1
In trigonometry identity
cot^2x = cos²x/sin²x
Csc²x = 1/sin²x
Substitute into the original expression
cos²x/sin²x - 1/sin²x
Find the LCM
(Cos²x-1)/sin²x .... *
Recall that sin²x+cos²x = 1
Sin²x = 1-cos²x
-sin²x = -1+cos²x
-sin²x = cos²x-1 .... **
Substitute ** into *
(Cos²x-1)/sin²x
-sin²x/sin²x
= -1 (RHS)
Therefore cot^2(x) - csc^2(x) = -1 (Proved!)
