Question #9385f

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Answer 1 · 2016-12-14T02:04:42

#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 · 2016-12-14T02:12:19

#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 · 2016-12-14T03:31:57

See Below

#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