Question

Can someone help me answer 61 and 63?

Answer 1

see explanation

Explanation:

using the trigonometric identities

• sin²x + cos²x = 1

• tan²x + 1 = sec²x

• cotx =
`(1)/(tanx)` and secx =
`(1)/(cosx)`

(61)

subtracting the 2 fractions

=
`(tan^2x-sec^2x)/(tanx)`

=
`(sec^2x-1-sec^2x)/(tanx)`

=
`(-1)/(tanx)`

= - cotx

(63)

adding the 2 fractions

=
`(cos^2x+(1+sinx)^2)/(cosx(1+sinx))`

=
`(cos^2x+1+2sinx+sin^2x)/(cosx(1+sinx))`

=
`(2(1+sinx))/(cosx(1+sinx))`

=
`(2)/(cosx)` = 2secx