# 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.Â  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

• sin2Â a + cos2Â a = 1
• 1+tan2Â aÂ  = sec2a
• cosec2Â a = 1 + cot2Â a

### Reciprocal Identities

• SinÂ Î¸ = $$\frac { 1 }{ CscÂ Î¸ }$$ or CscÂ Î¸ = $$\frac { 1 }{ SinÂ Î¸ }$$
• CosÂ Î¸ = $$\frac { 1 }{ SecÂ Î¸ }$$ or SecÂ Î¸ = $$\frac { 1 }{ CosÂ Î¸ }$$
• TanÂ Î¸ = $$\frac { 1 }{ CotÂ Î¸ }$$ or CotÂ Î¸ = $$\frac { 1 }{ TanÂ Î¸ }$$

### Opposite Angle Identities

• Sin (-Î¸) = â€“ SinÂ Î¸
• Cos (-Î¸) = CosÂ Î¸
• Tan (-Î¸) = â€“ TanÂ Î¸
• Cot (-Î¸) = â€“ CotÂ Î¸
• Sec (-Î¸) = SecÂ Î¸
• Csc (-Î¸) = -CscÂ Î¸

### Complementary Angles Identities

• Sin (90 â€“Â Î¸) = CosÂ Î¸
• Cos (90 â€“Â Î¸) = SinÂ Î¸
• Tan (90 â€“Â Î¸) = CotÂ Î¸
• Cot ( 90 â€“Â Î¸) = TanÂ Î¸
• Sec (90 â€“Â Î¸) = CscÂ Î¸
• Csc (90 â€“Â Î¸) = SecÂ Î¸

### Ratio Identities

• TanÂ Î¸ = $$\frac { SinÂ Î¸ }{ CosÂ Î¸ }$$
• CotÂ Î¸ = $$\frac { CosÂ Î¸ }{ SinÂ Î¸ }$$

### Angle Sum and Difference Identities

Consider two angles ,Â Î± and Î², the trigonometric sum and difference identities are as follows:

• sin(Î±+Î²)=sin(Î±).cos(Î²)+cos(Î±).sin(Î²)
• sin(Î±â€“Î²)=sinÎ±.cosÎ²â€“cosÎ±.sinÎ²
• cos(Î±+Î²)=cosÎ±.cosÎ²â€“sinÎ±.sinÎ²
• cos(Î±â€“Î²)=cosÎ±.cosÎ²+sinÎ±.sinÎ²
• tan(Î±+Î²) = $$\frac { tanÎ±+tanÎ² }{ 1-tanÎ±.tanÎ² }$$
• tan(Î±-Î²) =Â $$\frac { tanÎ±-tanÎ² }{ 1+tanÎ±.tanÎ² }$$

### Prove the following Trigonometric Identities

1. (1 – cos2Î¸) csc2Î¸Â  = Â 1?

Solution:

Let us consider L.H.S = Â (1 – cos2Î¸) csc2Î¸Â  andÂ  R.H.SÂ  = Â 1.

L.H.S = Â (1 – cos2Î¸) csc2Î¸

We know sin2Î¸Â +Â cos2Î¸Â  = Â 1,

sin2Î¸Â  =Â  1 –Â cos2Î¸

L.H.S = Â sin2Î¸Â â‹…Â csc2Î¸

We also know csc2Î¸ =Â  1/ sin2Î¸

L.H.S = Â sin2Î¸ â‹…Â  1 / sin2Î¸

L.H.S = 1

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

2. Prove tanÂ Î¸ sin Î¸ + cos Î¸ Â = Â sec Î¸

Solution:

Let L.H.S Â = Â tan Î¸ sin Î¸ + cos Î¸ Â and R.H.S = Â sec Î¸.

L.H.S = Â tan Î¸ sin Î¸ + cos Î¸

We know tanÎ¸ = $$\frac { SinÂ Î¸ }{ CosÂ Î¸ }$$

L.H.S = $$\frac { SinÂ Î¸ }{ CosÂ Î¸ }$$ â‹… sin Î¸ + cos Î¸

L.H.S =Â  Sin2 Î¸ / Cos Î¸+ cos Î¸

L.H.S = (sin2Î¸/cos Î¸) + (cos2Î¸/cosÎ¸)

L.H.S = (sin2Î¸Â + cos2Î¸) / cosÂ Î¸

L.H.S= $$\frac { 1}{ cos Î¸ }$$

L.H.S = 1 / cos Î¸

= Sec Î¸

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

3. Prove cot Î¸ + tan Î¸ Â = Â sec Î¸ csc Î¸?

Solution:

Let L.H.S Â = Â cot Î¸ + tan Î¸ and R.H.SÂ  = Â sec Î¸ csc Î¸.

L.H.S = Â cot Î¸ + tan Î¸

L.H.S =Â  $$\frac { CosÂ Î¸ }{ SinÂ Î¸ }$$ + $$\frac { SinÂ Î¸ }{ CosÂ Î¸ }$$

L.H.S = (cos2Î¸/sin Î¸ cosÂ Î¸) + (sin2Î¸/sinÂ Î¸Â cosÂ Î¸)

L.H.S = (cos2Î¸Â + sin2Î¸) / sinÂ Î¸ cosÂ Î¸

L.H.S = $$\frac { 1 }{ sin Î¸ cos Î¸ }$$

L.H.S = $$\frac { 1 }{ cos Î¸ }$$â‹… $$\frac { 1 }{ sin Î¸ }$$

L.H.S = Â sec Î¸ csc Î¸

L.H.S = R.H.S

4. Prove sec Î¸ âˆš(1 – sin2Î¸)Â  =Â  1?

Solution:

Let L.H.S Â = Â sec Î¸ âˆš(1 – sin2Î¸) Â and R.H.SÂ  = Â 1.

L.H.S = Â sec Î¸ âˆš(1 – sin2Î¸)

We know sin2Î¸Â +Â cos2Î¸Â  = Â 1,Â we have

cos2Î¸Â  =Â  1 – sin2Î¸

Then,

L.H.SÂ  = Â sec Î¸ âˆšcos2Î¸

L.H.S = Â sec Î¸ â‹… cos Î¸

L.H.S = Â sec Î¸ â‹… $$\frac { 1 }{ sec Î¸ }$$

L.H.S = $$\frac { sec Î¸ }{ sec Î¸ }$$

L.H.S = Â 1

L.H.S = Â R.H.S

5. Prove (1 – cos Î¸)(1 + cos Î¸)(1 + cot2Î¸) Â = Â 1

Solution:

Let L.H.SÂ  = Â (1 – cos Î¸)(1 + cos Î¸)(1 + cot2Î¸) Â = Â 1 and R.H.SÂ  = Â 1.

L.H.S = Â (1 – cos Î¸)(1 + cos Î¸)(1 + cot2Î¸)

L.H.S = Â (1 – cos2Î¸)(1 + cot2Î¸)

We know sin2Î¸Â +Â cos2Î¸Â  = Â 1,Â we have

sin2Î¸Â  =Â  1 –Â cos2Î¸

Then,

L.H.S = Â sin2Î¸Â â‹…Â (1 + cot2Î¸)

L.H.S = Â sin2Î¸Â Â +Â sin2Î¸Â â‹…Â cot2Î¸

L.H.S = Â sin2Î¸Â Â +Â sin2Î¸ â‹…Â (cos2Î¸/sin2Î¸)

L.H.S = Â sin2Î¸Â +Â cos2Î¸

L.H.S = Â 1

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

6. Prove tan4Î¸Â + tan2Î¸Â  =Â  sec4Î¸Â – sec2Î¸?

Solution:

Let L.H.S Â =Â Â tan4Î¸Â + tan2Î¸Â  and R.H.SÂ  =Â  sec4Î¸Â + sec2Î¸.

L.H.S =Â  tan4Î¸Â + tan2Î¸

L.H.S = Â tan2Î¸Â (tan2Î¸Â + 1)

We know that,

tan2Î¸Â  =Â  sec2Î¸ – 1

tan2Î¸ + 1Â  =Â Â sec2Î¸

Then,

L.H.S = Â (sec2Î¸Â – 1)(sec2Î¸)

L.H.SÂ  = Â sec4Î¸Â – sec2Î¸

7. Prove âˆš{(sec Î¸ â€“ 1)/(sec Î¸ + 1)} Â = Â cosec Î¸ – cot Î¸?

Solution:

Let L.H.SÂ  = âˆš{(sec Î¸ â€“ 1)/(sec Î¸ + 1)} and R.H.SÂ  = Â cosec Î¸ – cot Î¸

= âˆš{(sec Î¸ â€“ 1)/(sec Î¸ + 1)}

= âˆš[{(sec Î¸ – 1) (sec Î¸ – 1)}/{(sec Î¸ + 1) (sec Î¸ – 1)}]

= âˆš{(sec Î¸ – 1)2Â / (sec2Î¸Â – 1)}

= âˆš{(sec Î¸ – 1)2Â / tan2Î¸}

= (sec Î¸ â€“ 1)/tan Î¸

= Â (sec Î¸/tan Î¸) â€“ (1/tan Î¸)

= Â {(1/cos Î¸)/(sin Î¸/cos Î¸)} – cot Î¸

= {(1/cos Î¸)Â â‹…Â (cos Î¸/sin Î¸)} – cot Î¸

= Â (1/sin Î¸) – cot Î¸

= cosec Î¸ – cot Î¸

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

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