Question

1. If ø is an acute angle such that cosø =3/5, then

Sinø tanø -1
——————— =
2 tan^2ø

A) 16/625
B)1/36
C)3/160
D) 160/3

Answer 1

${\bold{\sf{➠\:cosø=3/5⟹sinø=4/5}}}$

${\bold{\sf{➠}}}$${\boxed{\sf{\:\frac{sinø.tanø-1}{2tan²ø}}}}$

${\bold{\sf{➠}}}$${\boxed{\sf{\:\frac{(4/5).((4/5)÷(3/5))-1}{2((4/5)÷(3/5))²}}}}$

${\bold{\sf{➠}}}$${\boxed{\sf{\:\frac{(4/5).(4/3)-1}{2(4/3)²}}}}$${\bold{\sf{➠}}}$${\boxed{\sf{\:\frac{(16/15)-1}{2(16/9)}}}}$

${\bold{\sf{➠}}}$${\boxed{\sf{\:\frac{1/15}{32/9}}}}$${\bold{\sf{➠}}}$${\boxed{\red{\sf{\:\frac{3}{160}}}}}$