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Help on Trigonometric Identities?

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csc^4 x – cot^4 x = csc^2 x + cot^2 x

(cosx + 1)/(sin^3 x)=(csc x)/(1 – cosx)

(sinθ + cosθ )^2+ (sinθ - cosθ )^2= 2
asked Oct 1, 2014 in TRIGONOMETRY by anonymous

3 Answers

0 votes

1) Left hand side identity = csc4x - cot4x

= (csc2x)2 - (cot2x)2

{ From pythagorean identities cot2x + 1 = csc2x }

= (cot2x + 1)2 - (cot2x)2

{ Apply the formula (a + b)= a2 + b2 + 2ab, In this case a = cotx , b = 1}

= cot4x + 1 + 2cot2x - cot4x

= 1 + 2cot2x

= 1 + cot2x + cot2x

= (1 + cot2x) + cot2x

= csc2x + cot2x

= Right hand side identity.

answered Oct 1, 2014 by david Expert
0 votes

2) Left hand side identity

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From pythagorean identities  sin2(x) + cos2(x) = 1

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= Right hand side identity.

answered Oct 1, 2014 by david Expert
0 votes

3) Left hand side identity = ( sinθ + cosθ )2 + ( sinθ - cosθ )2

{ Apply the formula (a + b)= a2 + b2 + 2ab }

= [ sin2θ + cos2θ + 2 sinθ cosθ ]+ [ sin2θ + cos2θ - 2 sinθ cosθ]

From pythagorean identities  sin2(θ) + cos2(θ) = 1

= [1 + 2 sinθ  cosθ] + [ 1 - 2 sinθ  cosθ]

= 1 + 2 sinθ  cosθ +  1 - 2 sinθ  cosθ

= 2

Right hand side identity.

answered Oct 1, 2014 by david Expert

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