Question

Evaluate without using tables:

`[(2cos 60°, -2sin 30°),(-tan45°, cos 0°)] [(cos 45°, cosec  30°),(sec 60°, sin 90°)]`

Answer 1

We know that,

cos 60° = `1/2`, sin 30° = `1/2`, tan 45° = 1, cos 0° = 1 

tan 45° = 1, cosec 30° = 2, sec 60° = 2, sin 90° = 1 

Now `[(2cos 60°, -2sin 30°),(-tan45°, cos 0°)] [(cos 45°, cosec 30°),(sec 60°, sin 90°)]`

= `[(2 xx 1/2, -2 xx 1/2),(-1, 1)][(1, 2),(2,1)]`

= `[(1, -1),(-1, 1)][(1, 2),(2, 1)]`

= `[(1 xx 1 + (-1) xx 2, 1 xx 2 + (-1) xx 1),(-1 xx 1 + 1 xx 2, -1 xx 2 + 1 xx 1)]`

= `[(1 - 2, 2 - 1),(-1 + 2, -2 + 1)]`

= `[(-1, 1),(1, -1)]`