Worksheet On Trigonometric Identities

Worksheet on Trigonometric Identities | Proving Trigonometric Identities Worksheets with Answers

Trigonometric Identities are useful when you are dealing with Trigonometric Functions in an Algebraic Expression. Usually, the Trig Identities involve certain functions of one or more angles. There are Several Identities involving the angle of a triangle and side length.  Check out all Fundamental Trigonometric Identities derived from Trigonometric Ratios using Worksheet on Trigonometric Identities. Practice the List of Trigonometric Identities, their derivation, and problems easily taking the help of the Trig Identities Worksheet with Answers.

List of Trigonometric Identities

There are several Trigonometric Identities that are used while solving Trigonometric Problems. Have a glance at the basic or fundamental trigonometric identities listed below and make your job simple. They are as follows

Pythagorean Identities

  • sina + cosa = 1
  • 1+tan2 a  = sec2a
  • coseca = 1 + cota

Reciprocal Identities

  • Sin θ = \(\frac { 1 }{ Csc θ } \) or Csc θ = \(\frac { 1 }{ Sin θ } \)
  • Cos θ = \(\frac { 1 }{ Sec θ } \) or Sec θ = \(\frac { 1 }{ Cos θ } \)
  • Tan θ = \(\frac { 1 }{ Cot θ } \) or Cot θ = \(\frac { 1 }{ Tan θ } \)

Opposite Angle Identities

  • Sin (-θ) = – Sin θ
  • Cos (-θ) = Cos θ
  • Tan (-θ) = – Tan θ
  • Cot (-θ) = – Cot θ
  • Sec (-θ) = Sec θ
  • Csc (-θ) = -Csc θ

Complementary Angles Identities

  • Sin (90 – θ) = Cos θ
  • Cos (90 – θ) = Sin θ
  • Tan (90 – θ) = Cot θ
  • Cot ( 90 – θ) = Tan θ
  • Sec (90 – θ) = Csc θ
  • Csc (90 – θ) = Sec θ

Ratio Identities

  • Tan θ = \(\frac { Sin θ }{ Cos θ } \)
  • Cot θ = \(\frac { Cos θ }{ Sin θ } \)

Angle Sum and Difference Identities

Consider two angles , α and β, the trigonometric sum and difference identities are as follows:

  • sin(α+β)=sin(α).cos(β)+cos(α).sin(β)
  • sin(α–β)=sinα.cosβ–cosα.sinβ
  • cos(α+β)=cosα.cosβ–sinα.sinβ
  • cos(α–β)=cosα.cosβ+sinα.sinβ
  • tan(α+β) = \(\frac { tanα+tanβ }{ 1-tanα.tanβ } \)
  • tan(α-β) = \(\frac { tanα-tanβ }{ 1+tanα.tanβ } \)

Prove the following Trigonometric Identities

1. (1 – cos2θ) csc2θ  =  1?

Solution:

Let us consider L.H.S =  (1 – cos2θ) csc2θ  and  R.H.S  =  1.

L.H.S =  (1 – cos2θ) csc2θ

We know sin2θ + cos2θ  =  1,

sin2θ  =  1 – cos2θ

L.H.S =  sin2θ ⋅ csc2θ

We also know csc2θ =  1/ sin2θ

L.H.S =  sin2θ ⋅  1 / sin2θ

L.H.S = 1

Hence Proved, L.H.S = R.H.S


2. Prove tan θ sin θ + cos θ  =  sec θ

Solution:

Let L.H.S  =  tan θ sin θ + cos θ  and R.H.S =  sec θ.

L.H.S =  tan θ sin θ + cos θ

We know tanθ = \(\frac { Sin θ }{ Cos θ } \)

L.H.S = \(\frac { Sin θ }{ Cos θ } \) ⋅ sin θ + cos θ

L.H.S =  Sin2 θ / Cos θ+ cos θ

L.H.S = (sin2θ/cos θ) + (cos2θ/cosθ)

L.H.S = (sin2θ + cos2θ) / cos θ

L.H.S= \(\frac { 1}{ cos θ } \)

L.H.S = 1 / cos θ

= Sec θ

Therefore, L.H.S = R.H.S


3. Prove cot θ + tan θ  =  sec θ csc θ?

Solution:

Let L.H.S  =  cot θ + tan θ and R.H.S  =  sec θ csc θ.

L.H.S =  cot θ + tan θ

L.H.S =  \(\frac { Cos θ }{ Sin θ } \) + \(\frac { Sin θ }{ Cos θ } \)

L.H.S = (cos2θ/sin θ cos θ) + (sin2θ/sin θ cos θ)

L.H.S = (cos2θ + sin2θ) / sin θ cos θ

L.H.S = \(\frac { 1 }{ sin θ cos θ } \)

L.H.S = \(\frac { 1 }{ cos θ } \)⋅ \(\frac { 1 }{ sin θ } \)

L.H.S =  sec θ csc θ

L.H.S = R.H.S


4. Prove sec θ √(1 – sin2θ)  =  1?

Solution:

Let L.H.S  =  sec θ √(1 – sin2θ)  and R.H.S  =  1.

L.H.S =  sec θ √(1 – sin2θ)

We know sin2θ + cos2θ  =  1, we have

cos2θ  =  1 – sin2θ

Then,

L.H.S  =  sec θ √cos2θ

L.H.S =  sec θ ⋅ cos θ

L.H.S =  sec θ ⋅ \(\frac { 1 }{ sec θ } \)

L.H.S = \(\frac { sec θ }{ sec θ } \)

L.H.S =  1

L.H.S =  R.H.S


5. Prove (1 – cos θ)(1 + cos θ)(1 + cot2θ)  =  1

Solution:

Let L.H.S  =  (1 – cos θ)(1 + cos θ)(1 + cot2θ)  =  1 and R.H.S  =  1.

L.H.S =  (1 – cos θ)(1 + cos θ)(1 + cot2θ)

L.H.S =  (1 – cos2θ)(1 + cot2θ)

We know sin2θ + cos2θ  =  1, we have

sin2θ  =  1 – cos2θ

Then,

L.H.S =  sin2θ ⋅ (1 + cot2θ)

L.H.S =  sin2θ  + sin2θ ⋅ cot2θ

L.H.S =  sin2θ  + sin2θ ⋅ (cos2θ/sin2θ)

L.H.S =  sin2θ + cos2θ

L.H.S =  1

Therefore, L.H.S = R.H.S


6. Prove tan4θ + tan2θ  =  sec4θ – sec2θ?

Solution:

Let L.H.S  =  tan4θ + tan2θ  and R.H.S  =  sec4θ + sec2θ.

L.H.S =  tan4θ + tan2θ

L.H.S =  tan2θ (tan2θ + 1)

We know that,

tan2θ  =  sec2θ – 1

tan2θ + 1  =  sec2θ

Then,

L.H.S =  (sec2θ – 1)(sec2θ)

L.H.S  =  sec4θ – sec2θ


7. Prove √{(sec θ – 1)/(sec θ + 1)}  =  cosec θ – cot θ?

Solution:

Let L.H.S  = √{(sec θ – 1)/(sec θ + 1)} and R.H.S  =  cosec θ – cot θ

= √{(sec θ – 1)/(sec θ + 1)}

= √[{(sec θ – 1) (sec θ – 1)}/{(sec θ + 1) (sec θ – 1)}]

= √{(sec θ – 1)/ (sec2θ – 1)}

= √{(sec θ – 1)/ tan2θ}

= (sec θ – 1)/tan θ

=  (sec θ/tan θ) – (1/tan θ)

=  {(1/cos θ)/(sin θ/cos θ)} – cot θ

= {(1/cos θ) ⋅ (cos θ/sin θ)} – cot θ

=  (1/sin θ) – cot θ

= cosec θ – cot θ

Therefore, L.H.S = R.H.S


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