Find the exact values of the six trigonometric functions for angle e in standard position if a point with the coordinates (-6, 6) lies on its terminal side. 13 note: a value such as = can be entered as sr3/2.

tan e = -1

cot e = -1

sin e = √2/2

cosec e = √2

cos e = -√2/2

sec e = -√2.

Step-by-step explanation:

6/6-  is the tangent of e  so tan e = -1.

cot e = 1/tan e = -1.

The hypotenuse  of the triangle containing angle e = √(-6)^2 + (6)^2 ( By the pythagoras theorem) and = √72 =  6√2.

Therefore sin e =   6/6√2

= 1/√2

= √2/2

cosec e  = 1 ./ sin e = √2.

cos e = -6 / 6√2

= -√2/2.

sec e = 1/cos e = -√2.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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