Problems on Trigonometric Identities

Here we will prove the problems on trigonometric identities. In an identity there are two sides of the equation, one side is known as ‘left hand side’ and the other side is known as ‘right hand side’ and to prove the identity we need to use logical steps showing that one side of the equation ends up with the other side of the equation.

Proving the problems on trigonometric identities:

1. (1 - sin A)/(1 + sin A) = (sec A - tan A)2

Solution:

L.H.S = (1 - sin A)/(1 + sin A)

= (1 - sin A)2/(1 - sin A) (1 + sin A),[Multiply both numerator and denominator by (1 - sin A)



= (1 - sin A)2/(1 - sin2 A)

= (1 - sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 - sin2 θ]

= {(1 - sin A)/cos A}2

= (1/cos A - sin A/cos A)2

= (sec A – tan A)2 = R.H.S. Proved.


2. Prove that, √{(sec θ – 1)/(sec θ + 1)} = cosec θ - cot θ.

Solution:

L.H.S.= √{(sec θ – 1)/(sec θ + 1)}

= √[{(sec θ - 1) (sec θ - 1)}/{(sec θ + 1) (sec θ - 1)}]; [multiplying numerator and denominator by (sec θ - l) under radical sign]

= √{(sec θ - 1)2/(sec2 θ - 1)}

=√{(sec θ -1)2/tan2 θ}; [since, sec2 θ = 1 + tan2 θ ⇒ sec2 θ - 1 = tan2 θ]

= (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 = R.H.S. Proved.


3. tan4 θ + tan2 θ = sec4 θ - sec2 θ

Solution:

L.H.S = tan4 θ + tan2 θ

= tan2 θ (tan2 θ + 1)

= (sec2 θ - 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]

= (sec2 θ - 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]

= sec4 θ - sec2 θ = R.H.S. Proved.

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More problems on trigonometric identities are shown where one side of the identity ends up with the other side.

4. . cos θ/(1 - tan θ) + sin θ/(1 - cot θ) = sin θ + cos θ

Solution:

L.H.S = cos θ/(1 - tan θ) + sin θ/(1 - cot θ)

= cos θ/{1 - (sin θ/cos θ)} + sin θ/{1 - (cos θ/sin θ)}

= cos θ/{(cos θ - sin θ)/cos θ} + sin θ/{(sin θ - cos θ/sin θ)}

= cos2 θ/(cos θ - sin θ) + sin2 θ/(cos θ - sin θ)

= (cos2 θ - sin2 θ)/(cos θ - sin θ)

= [(cos θ + sin θ)(cos θ - sin θ)]/(cos θ - sin θ)

= (cos θ + sin θ) = R.H.S. Proved.


5. Show that, 1/(csc A - cot A) - 1/sin A = 1/sin A - 1/(csc A + cot A)

Solution:

We have,

1/(csc A - cot A) + 1/(csc A + cot A)

= (csc A + cot A + csc A - cot A)/(csc2 A - cot2 A)

= (2 csc A)/1; [since, csc2 A = 1 + cot2 A ⇒ csc2A - cot2 A = 1]

= 2/sin A; [since, csc A = 1/sin A]

Therefore,

1/(csc A - cot A) + 1/(csc A + cot A) = 2/sin A

⇒ 1/(csc A - cot A) + 1/(csc A + cot A) = 1/sin A + 1/sin A

Therefore, 1/(csc A - cot A) - 1/sin A = 1/sin A - 1/(csc A + cot A) Proved.


6. (tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ

Solution:

L.H.S = (tan θ + sec θ - 1)/(tan θ - sec θ + 1)

= [(tan θ + sec θ) - (sec2 θ - tan2 θ)]/(tan θ - sec θ + 1), [Since, sec2 θ - tan2 θ = 1]

= {(tan θ + sec θ) - (sec θ + tan θ) (sec θ - tan θ)}/(tan θ - sec θ + 1)

= {(tan θ + sec θ) (1 - sec θ + tan θ)}/(tan θ - sec θ + 1)

= {(tan θ + sec θ) (tan θ - sec θ + 1)}/(tan θ - sec θ + 1)

= tan θ + sec θ

= (sin θ/cos θ) + (1/cos θ)

= (sin θ + 1)/cos θ

= (1 + sin θ)/cos θ = R.H.S. Proved.


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● Trigonometric Functions





10th Grade Math

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