SS1 Maths sin cos tan equatio

1)
sin^2 B + cos^2 B = 1
(-2/3)^2 + cos^2 B = 1
cos^2 B = 1 - 4/9 = 5/9
cos B = √5 / 3 or - √5 /3 (rejected since cos B > 0)
the value of cos B - tan B
= cos B - (sinB)/cosB
= √5 /3 - (-2/3)/ (√5 /3)
= √5 /3 + 2√5 /5
= 11√5 /15
2)
tan (180 + B) + tan (90 - B)
= tan B + 1/tan B

2010-06-07 13:47:44 補充：
= (sin B)/cosB + (cosB)/sinB

= [(sinB)^2 + (cosB)^2] / (cosB sinB)

= 1 / (cosB sinB)

= 2 / sin2B

2010-06-07 14:12:43 補充：
樓主品學兼憂 :

For Q1 :

My answer 11√5 /15

= 11√5 / (3 * 5)

= 11 / 3√5 = Given answer.


For Q2 :

I think the given answer is wrong ,

Set B = 45

tan (180 + B) + tan (90 - B)

= tan 225 + tan 45

= 1 + 1

= 2

But the given answer 1 / (sinB tanB)

= 1 / (sin45 tan45)

= √2 not = 2

2010-06-07 14:12:51 補充：
The answer should be 1 / (cosB sinB)

1 / (cos45 sin45) = 1 / (√2/2 * √2/2) = 2

2010-06-07 16:51:41 補充：
= sinB / cosB + cosB / sinB

= [(sinB)^2 + (cosB)^2] / (sinB cosB)

= 1 / (sinB cosB)

or

= 2 / 2(sinB cosB)

= 2 / sin2B

1.
sin B = -2/3
(sin B) ^2 = 4/9
1 - (cos B) ^2 = 4/9
(cos B) ^2 = 5/9
cos B = (squr 5)/3
[cos B > 0]
-------------
cos B - tan B
=cos B - sin B/cos B
=(cos B) ^2/cos B - sin B/cos B
=((cos B) ^2 - sin B)/cos B
=(5/9 - (-2/3)) / (squr 5)/3
=(7/9)/((squr 5)/3)
=7/3(squr 5)
-------------
2.
tan (180 + B) = tan B
--------------
tan (90 - B)
=sin (90 - B) / cos (90 - B)
=cos B/sin B
=1/tan B
---------------
tan (180 + B) + tan (90 - B)
=tan B + 1/tan B
=sin B/cos B + cos B/sin B
=(sin B) ^2/sin B cos B + (cos B) ^2/ sin B cos B
=((sin B) ^2 + (cos B) ^2)/sin B cos B
=1/sin B cos B

2010-06-07 13:57:05 補充：
第一題計錯...

=(5/9 - (-2/3)) / (squr 5)/3--->=(5/9 - (-2/3)) / (squr 5)/3
=(7/9)/((squr 5)/3)------------>=(11/9)/((squr 5)/3)
=7/3(squr 5)------------------->=11/3(squr 5)


11/3(squr 5)
=11/3(squr 5) x squr 5/squr 5
=11(squr 5) /15