Question

Question #9385f

Answer 1

`cscB/sinB - cotB/tanB = 1`

Use the identities `tantheta = sin theta/costheta`, `cottheta = 1/tantheta = 1/(sin theta/costheta) = costheta/sintheta` and `csctheta = 1/sintheta`.

`=>(1/sinB)/sinB - (cosB/sinB)/(sinB/cosB) = 1`

`=>1/sin^2B - cos^2B/sin^2B = 1`

`=> (1- cos^2B)/sin^2B = 1`

Use the identity `cos^2theta + sin^2theta = 1`.

`=> sin^2B/sin^2B = 1`

`=> 1 = 1`

`LHS = RHS`

Identity proved!

Hopefully this helps!

Answer 2

`cscB/sinB-cotB/tanB=1`

Use the identity `sinB=1/cscB`

`frac(cscB)(1/cscB)-cotB/tanB-1`

`cscB*cscB/1-cotB/tanB=1`

`csc^2B-cotB/tanB=1`

Use the identity `tanB=1/cotB`

`csc^2B-frac(cotB)(1/cotB)=1`

`csc^2B-cotB*cotB/1=1`

`csc^2B-cot^2B=1`

Use the Pythagorean identity `1+cot^2B=csc^2B`

`1+cot^2B-cot^2B=1`

`1=1`

Answer 3

See Below

Explanation:

`csc B/ sinB - cot B/ tan B = 1`

Left Hand Side:

`=csc B/ sinB - cot B / tan B`

`= csc B*1/sin B - cot B* 1/ tan B`

`=csc B csc B - cot B cot B`

`= csc^2 B - cot^2 B`

`=1` -->Use the identity `cot^2 B +1 = csc^2 B` and subtract `cot^2 B` from both sides to isolate 1.

`:.=` Right Hand Side