PROVING TRIGONOMETRIC IDENTITIES

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Abbreviations used in the problems :

* L.H.S -----> Left hand side

* R.H.S -----> Right hand side

Problem 1 :

Prove the following trigonometric identities.

(i) cotθ + tanθ  =  secθ cscθ

Solution :

cotθ  =  cosθ/sinθ

tanθ  =  sinθ/cosθ

L.H.S :

The reciprocal formula for 1/sin θ is cosec θ and 1/cos θ is sec θ.

Hence it is proved.

(ii) tan4θ + tan2θ  =  sec4θ − sec2θ

Solution :

L.H.S :

tan4θ + tan2θ  =  (tan2θ)2tan2θ

  =  tan2θ(tan2θ + 1)

  =  tan2θ(sec2θ)

  =  (sec2θ - 1)(sec2θ)

  =  sec4θ - sec2θ

R.H.S

Hence it is proved.

Problem 2 :

Prove the following identities.

(i)  (1 - tan2 θ)/(cot2 θ - 1)  =  tan2 θ

Solution :

L.H.S :

  =  (1 - tan2 θ) / (cot2 θ - 1)

Formula for tan θ and cot θ 

cot θ  =  cos θ / sin θ

tan θ  =  sin θ / cos θ

  =  tan2 θ

R.H.S

Hence it is proved.

(ii) cos θ/(1 + sin θ)  =  sec θ + tan θ

Solution :

L.H.S :

  =  cos θ/(1 + sin θ) 

By multiplying the conjugate of denominator, we get

  =  [cos θ/(1 + sin θ)] [(1 - sin θ)/(1 - sin θ)]

  =  [cos θ(1 - sin θ)/(1 - sin θ)(1 + sin θ)]

  =  [cos θ(1 - sin θ)/(1 - sinθ)]

  =  [cos θ(1 - sin θ)/cosθ]

  =  (1 - sin θ)/cos θ

  =  (1/cos θ) - (sin θ/cos θ)

  =  sec θ - tan θ

R.H.S

Problem 3 :

Prove the following identities.

(i)  √[(1 + sin θ)/(1 - sin θ)]  =  sec θ + tan θ

Solution :

L.H.S :

  =  (1/cos θ) + (sin θ/cos θ)

  =  sec θ + tan θ

 R.H.S

Hence proved.

(ii)  [√(1 + sin θ)/(1 - sin θ)] + [√(1 - sin θ)/(1 + sin θ)]  = 2 sec θ

Solution :

L.H.S : 

   =   [√(1 + sin θ)/(1 - sin θ)] + [√(1 - sin θ)/(1 + sin θ)]

  =  2/cos θ

  =  2 sec θ

R.H.S 

Hence proved .

Problem 4 :

Prove the following identities.

(i) secθ = tan6 θ + 3 tan2 θ sec2 θ + 1

Solution :

R.H.S : 

tan6 θ + 3 tan2 θ sec2 θ + 1

  =  (tan2 θ)3 + 3 (tan2 θ) (sec2 θ) + 1

(a + b)3  =  a3 + b3 + 3ab (a + b) 

  =  (tan2 θ)3 + 3 (tan2 θ) 1 (tan2 θ + 1) + 13

=  (tan2 θ + 1)3

=  (sec2 θ)3

=  sec6 θ

R.H.S

Hence proved.

(ii) (sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2

Solution :

L.H.S :

  =  (sin θ + sec θ)2 + (cos θ + cosec θ)2

  =  sin2 θ + sec2 θ + 2 sin θ sec θ + cos2 θ + cosec2 θ + 2 cos θ cosec θ

  =  (sin2 θ + cos2 θ) + sec2 θ  + cosec2 θ + 2 (sin θ sec θ + cos θ cosec θ)

  =  1 + sec2 θ  + cosec2 θ + 2 [(sin θ/cos θ) + (cos θ/sin θ)]

  =  1 + sec2 θ  + cosec2 θ + 2 [(sin2θ + cos2θ)/cos θ sin θ)]

  =  1 + sec2 θ  + cosec2 θ + 2 (1/cos θ sin θ)

  =  1 + sec2 θ  + cosec2 θ + 2 secθ cosec θ 

  =  1 + (sec θ  + cosec θ)2

R.H.S

Hence proved.

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