mohon dibantu ya besok dikumpulkan

³1.  krna 270 < A < 360  --> <A di kuadran IV
a. cos A = 3/5
sin A = √(1- cos² A) = √(1- (3/5)² = √(16/25) = +- 4/5
krn A di KD IV maka
sin A  = - 4/5
tan A = sin A/cos A = - 4/3
sec A = 1/cos A = 5/3
csc A = 1/sin A = - 3/4
cot A = 1/tan A = - 3/4

b. <A tumpul , <A di Kuadran II
sin A = √3/3
cos A = √(1 - (√3/ 3)² = √(1 - 1/3)=+_ √(2/3)
A di KD II, cos A < 0 --> cos  A = - 1/3 √6
tan A = sin A /cos A = - 1/2 √2
sec A = 1/cos A = - 3/√6= - 1/2 √6
csc A = 1/sin A = 3/√3  = √3

2. Buktikan sin A . sec A = tan A
ruas kiri sin A . sec A =
= sin A . (1/ cos A)
= sin A / cos A
= tan A = ruas kanan terbukti

4. sec A (1 - sin² A) = cos A
ruas kiri sec A (1 - sin² A) =
= 1/cos A ( cos ² A)
= cos A

5. tan A + (cos A) /(1 +sin A) = sec a
ruas kiri
tan A + (cos A)/ (1 + sin A) =
= sin A /cos A  + cos A /(1 +sin A)
= (sin A(1+sin A) + cos A . cos A)/ cos A (1 + sin A)
= ( sin A + sin² A + cos² A) / (cos A (1 + sin A))
= (sin A + 1) /(cos A (1+sin A)
= 1/cos A
= sec A

6.(1- sin A)/(1 + sin A) =(sec A - tan A)²
ruas kiri
(1 - sin A)/(1 +sin A) =
= (1 -sin A)(1- sin A) / (1+sin A)(1 -sin A)
= (1 -sin A)² / ( 1 - sin² A)
= (1 - sin A)² / cos² A
= {(1 - sin a) / cos A}²
= (1/cos A - sin A/cos A}²
= (sec A - tan A)²