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Please help me in verifying Trig Identities:

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1.( 2-cos^2x)/sinx = cscx+sinx

2. (1-cscx)/cscx=sinx-1/sinx+1

3. csc^2x-cos^2x-sin^2x=cot^2x?

asked Oct 31, 2014 in TRIGONOMETRY by anonymous

3 Answers

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1) (2 - cos2x)/sinx = cscx + sinx

Left hand side identity = (2 - cos2x)/sinx

[Recall the Pythagorean identities sin2x + cos2x = 1]

= [2 - (1 - sin2x)]/sinx

= (2 - 1 + sin2x)/sinx

= (1 + sin2x)/sinx

= 1/sinx + sin2x/sinx

= cscx + sinx

= Right hand side identity.

answered Oct 31, 2014 by david Expert
0 votes

3) csc2x - cos2x - sin2x = cot2 x

Left hand side identity = csc2x - cos2x - sin2x

{Recall the Pythagorean identities sin2x + cos2x = 1 and reciprocal identity of sinx is cscx}

= (1/sin2x) - ( 1 - sin2x ) - sin2x

= (1/sin2x) - 1 + sin2x  - sin2x

= (1/sin2x) - 1

= (1 - sin2x)/sin2x

= cos2x/sin2x

= cot2x

= Right hand side identity.

answered Oct 31, 2014 by david Expert
0 votes

2) Assume the identity ( 1 - cscx)/(1 + cscx) = (sinx - 1)/(sinx + 1)

Left hand side identity = (1 - cscx)/(1 + cscx)

{Reciprocal identity of cscx is sinx }

= [ 1 - (1/sinx)]/[1 + (1/sinx)]

= [(sinx - 1)/sinx]/[(sinx + 1)/sinx]

= (sinx - 1)/(sinx + 1)

= Right hand side identity.

answered Oct 31, 2014 by david Expert

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