Question

Q 1 PLEASE HELP ME FIGURE THIS OUT 1.0

Answer 1

Answer:
`(2√(2)-√(3))/(2)`

Step-by-step explanation (using the Unit Circle):

csc
`(\pi)/(4)` - cos
`(\pi)/(6)`

csc =
`(1)/(sin)`

sin
`(\pi)/(4)` =
`(√(2))/(2)`

→ csc
`(\pi)/(4)` =
`(2)/(√(2))`=
`{√(2)}`

cos
`(\pi)/(6)` =
`(√(3))/(2)`

csc
`(\pi)/(4)` - cos
`(\pi)/(6)`

=
`{√(2)}` -
`(√(3))/(2)`

=
`{√(2)}*((2)/(2))` -
`(√(3))/(2)`

=
`(2√(2))/(2)` -
`(√(3))/(2)`

=
`(2√(2)-√(3))/(2)`

Step-by-step explanation (using the special triangles):

`(\pi)/(4)` = 45°

a 45°-45°-90° triangle has sides with proportions of: 1 - 1 - √2

csc =
`(hypotenuse)/(opposite)` =
`\frac{√(2)} {1} =√(2)`

`(\pi)/(6)` = 30°

a 30°-60°-90° triangle has sides with proportions of: 1 - √3 - 2

cos =
`(adjacent)/(hypotenuse)` =
`(√(3))/(2)`

csc
`(\pi)/(4)` - cos
`(\pi)/(6)` =
`\frac{2√(2)-√(3)} {2}`

*****************************************************************************************

Answer: secθ

Explanation:

cosθ csc²θ tan²θ

= cosθ *
`(1)/(sin^(2)\theta)` *
`\frac{sin^(2)\theta} {cos^(2)\theta}`

=
`(1)/(cos\theta)`

= secθ