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3.Trigonometrical Ratios, Functions and Identities
easy
If $x = \sin {130^o}\,\cos {80^o},\,\,y = \sin \,{80^o}\,\cos \,{130^o},\,\,z = 1 + xy,$which one of the following is true
A
$x > 0,\,\,y > 0,\,\,z > 0$
B
$x > 0,\,\,y < 0,\,\,0 < z < 1$
C
$x > 0,\,\,y < 0,\,\,z > 1$
D
$x < 0,\,\,y < 0,\,0 < z < 1$
Solution
(b) $x = \sin {130^o}\cos {80^o},$
$y = \sin {80^o}\cos {130^o}$
==> $x = \cos {40^o}\cos {80^o},\,\,\,y = – \sin {80^o}\sin {40^o}$
So, $x > 0$ and $y < 0$ and $xy < 0$
Now, $z = 1 + xy$ ==> $0 < z < 1$.
Standard 11
Mathematics
