F.4 Amaths兩條

As follows:

圖片參考：http://i190.photobucket.com/albums/z177/ansonluizzzz/1-5.jpg

( 1 ) sin^2@ / ( 1 + 2 cos^2 @ ) = 3 / 19
( 1 - cos^2 @ ) / ( 1 + 2 cos^2 @ ) = 3 / 19
25cos^2 @ = 16
cos^2 @ = 16 / 25
cos @ = - 4 / 5 since 180* < @ < 270*
tan @ = 3 / 4 and sec @ = - 5 / 4
So,
tan @ / ( 1 + sec @ )
= ( 3 / 4 ) / ( 1 - 5 / 4 )
= -3
2) sec @ ( sec @ - tan @ ) = x
sec^2 @ - sin@ sec^2 @ = x
sec^2 @ ( 1 - sin @ ) = x
( 1 - sin @ ) / ( 1 - sin^2 @ ) = x
1 / ( 1 + sin @ ) = x
sin @ = ( 1 - x ) / x or ( x - 1 ) / x

Ans:
1) - 3