prove (tanB+cotB)/tan B= csc^2 B?

Answer 1

LHS
[sinB / cosB + cosB /sinB ] (cosB / sinB)
[(sin²B + cos²B) / (sinB cosB) ] (cos B / sin B)
[ (1) / (sin B cos B) ] (cos B / sin B)
1 / sin ² B
cosec ² B

RHS
cosec ² B

LHS = RHS

Answer 2

Tan B = sinB/cosB
cot B = cosB/sinB
so TanB+CotB = sinB/cosB +cosB/sinB

=[sin^2(B) +cos^(2)B]/sinBcosBtanB
=1/[sinBcosB(sinB/cosB)]
=1/sin^2(B)=csc^2(B)

Answer 3

l.s={(sinb/cosb)+(cosb+sinb)}/(sinb/cosb)

get a common denominator by multiplying the two denominators to get
{(sin^2b+cos^2b)/cosbsinb}/(sinb/cosb)

notice how sin^2b + cos^2b turns to 1

---> (1/cosbsinb )/sinbcosb

THen you reduce it by dividing by the denominator to get (flip and multiply the denomintor (sinbcosb)
you will get 1/sin^2b ----> which is csc^2B

Sometimes you have to work on both sides but this one, you only had to do the left side ..start with the side that has more stuff and simplify it as much as you can.

Good luck with trig...it takes a soldier to ace it

Answer 4

prove (tanB+cotB)/tan B= csc^2 B

tanB/tanB + cotB/tanB = csc^2B
1 + cscBcosB/secBsinB = csc^2B
1 + [(cscB/secB)(cosB/sinB)] = csc^2B
1 + [cotBcotB] = csc^2B
1 + cot^2B = csc^2B
csc^2B = csc^2B ANS

teddy boy

Answer 5

all lets start out with left side.

(sinb/cosb + cosb/sinb) / tanb
=[(sin^2b + cos^2b)/ sinbcosb] / (sinb/cos) add the numerator by getting common denominator. and the tanb = sinb/cosb for the denominator
so now
=[ (sin^2b + cos^2b) * Cosb] / [cosb sin^2b]

= [sin^2b+cos^2b] / sin^2b

= 1/sin^2b
= csc^2b

qed

Answer 6

tanB = sinB/cosB
cotB = cosB/sinB
cscB = 1/sinB
sin²B + cos²B = 1

LHS
= (tanB + cotB) / tanB
= (sinB/cosB cosB/sinB) / (sinB/cosB)
= ((sin²B + cos²B) / (sinBcosB)) / (sinB/cosB)
= cosB / (sinB(sinBcosB))
= 1/sin²B
= csc²B
= RHS

Answer 7

note that:
tanB = sinB/cosB
cotB = cosB/sinB
csc²B = 1 / sin²B
sin²B + cos²B = 1

Left side:

=(sinB/cosB + cosB/sinB) / (sinB/cosB)
=[(sin²B + cos²B) / sinBcosB] / (sinB/cosB)
=[1 / sinBcosB] [cosB/sinB]
=1 / sin²B

Right side:

= 1 / sin²B

LS = RS, therefore the identity is true.

Answer 8

that isn't even a math equation!