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
- sin2 a + cos2 a = 1
- 1+tan2 a = sec2a
- cosec2 a = 1 + cot2 a
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)2 / (sec2θ – 1)}
= √{(sec θ – 1)2 / 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