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Class 10 RD Sharma Solutions – Chapter 6 Trigonometric Identities – Exercise 6.1 | Set 2

  • Last Updated : 30 Apr, 2021
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Prove the following trigonometric identities:

Question 29. \frac{1+secθ}{secθ}=\frac{sin^2θ}{1-cosθ}      

Solution:

We have,

L.H.S. = \frac{1+secθ}{secθ}

\frac{1+\frac{1}{cosθ}}{\frac{1}{cosθ}}

\frac{cosθ+1}{cosθ}×{\frac{cosθ}{1}}



= 1 + cos θ

\frac{(1+cosθ)(1-cosθ)}{1-cosθ}

\frac{1-cos^2θ}{1-cosθ}

\frac{sin^2θ}{1-cosθ}

= R.H.S.

Hence proved.

Question 30. \frac{tanθ}{1-cotθ}+\frac{cotθ}{1-tanθ}=1+tanθ+cotθ

Solution:

We have,



L.H.S. = \frac{tanθ}{1-cotθ}+\frac{cotθ}{1-tanθ}

\frac{tanθ}{1-\frac{1}{tanθ}}+\frac{\frac{1}{tanθ}}{1-tanθ}      

\frac{1}{1-tanθ}[\frac{1}{tanθ}-tan^2θ]

\frac{1}{1-tanθ}[\frac{1-tan^3θ}{tanθ}]

\frac{1}{1-tanθ}\left[\frac{(1-tanθ)(1+tanθ+tan^2θ)}{tanθ}\right]

\frac{1+tanθ+tan^2θ}{tanθ}

\frac{1}{tanθ}+\frac{tanθ}{tanθ}+\frac{tan^2θ}{tanθ}

= 1 + tanθ + cotθ

= R.H.S.

Hence proved.



Question 31. sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1

Solution: 

We know,

sec2 θ − tan2 θ = 1

On cubing both sides, we get,

=> (sec2θ − tan2θ)3 = 1

=> sec6 θ − tan6 θ − 3sec2 θ tan2 θ(sec2 θ − tan2 θ) = 1

=> sec6 θ − tan6 θ − 3sec2 θ tan2 θ = 1

=> sec6 θ = tan6 θ + 3sec2 θ tan2 θ + 1

Hence proved.

Question 32. cosec6 θ = cot6 θ + 3cot2 θ cosec2 θ + 1

Solution:

We know,

cosec2 θ − cot2 θ = 1

On cubing both sides,

=> (cosec2 θ − cot2 θ)3 = 1

=> cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ (cosec2 θ − cot2 θ) = 1

=> cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ = 1

=> cosec6 θ = cot6 θ + 3 cosec2 θ cot2 θ + 1

Hence proved.

Question 33. \frac{(1+tan^2θ)cotθ}{cosec^2θ}=tanθ

Solution:

We have,



L.H.S. = \frac{(1+tan^2θ)cotθ}{cosec^2θ}

\frac{sec^2θcotθ}{cosec^2θ}

\frac{sin^2θcosθ}{cos^2θsinθ}

= sin θ/cos θ

= tan θ

= R.H.S.

Hence proved.

Question 34. \frac{1+cosA}{sin^2A}=\frac{1}{1-cosA}

Solution:

We have,

L.H.S. = \frac{1+cosA}{sin^2A}



\frac{(1+cosA)(1-cosA)}{(sin^2A)(1-cosA)}

\frac{1-cos^2A}{(sin^2A)(1-cosA)}

\frac{sin^2A}{(sin^2A)(1-cosA)}

\frac{1}{1-cosA}

= R.H.S.

Hence proved.

Question 35. \frac{secA-tanA}{secA+tanA}=\frac{cos^2A}{(1+sinA)^2}

Solution:

We have,

L.H.S. = \frac{secA-tanA}{secA+tanA}

\frac{(secA-tanA)(secA+tanA)}{(secA+tanA)(secA+tanA)}



\frac{(sec^2A-tan^2A)}{(secA+tanA)^2}

\frac{1}{sec^2A+tan^2A+2secAtanA}

\frac{1}{\frac{1}{cos^2A}+\frac{sin^2A}{cos^2A}+\frac{2sinA}{cos^2A}}

\frac{cos^2A}{1+sin^2A+2sinA}

\frac{cos^2A}{(1+sinA)^2}

= R.H.S.

Hence proved.

Question 36. \frac{1+cosA}{sinA}=\frac{sinA}{1-cosA}

Solution:

We have,

L.H.S. = \frac{1+cosA}{sinA}



\frac{(1+cosA)(1-cosA)}{sinA(1-cosA)}

\frac{1-cos^2A}{sinA(1-cosA)}

\frac{sin^2A}{sinA(1-cosA)}

\frac{sinA}{1-cosA}

= R.H.S.

Hence proved.

Question 37. (i) \sqrt{\frac{1+sinA}{1-sinA}}=secA+tanA

Solution:

We have,

L.H.S. = \sqrt{\frac{1+sinA}{1-sinA}}

\sqrt{\frac{(1+sinA)(1+sinA)}{(1-sinA)(1+sinA)}}      



\sqrt{\frac{(1+sinA)^2}{1-sin^2A}}

\frac{1+sinA}{cosA}

\frac{1}{cosA}+\frac{sinA}{cosA}

= sec A + tan A

= R.H.S.

Hence proved.

(ii) \sqrt{\frac{1-cosA}{1+cosA}}+\sqrt{\frac{1+cosA}{1-cosA}}=2cosecA

Solution:

We have,

L.H.S. = \sqrt{\frac{1-cosA}{1+cosA}}+\sqrt{\frac{1+cosA}{1-cosA}}

\frac{1-cosA+1+cosA}{\sqrt{1-cos^2A}}

\frac{2}{sinA}

= 2 cosec A

= R.H.S.

Hence proved.

Question 38. (i) \sqrt{\frac{secθ-1}{secθ+1}}+\sqrt{\frac{secθ+1}{secθ-1}}=2cosecθ

Solution:

We have, 

L.H.S. = \sqrt{\frac{secθ-1}{secθ+1}}+\sqrt{\frac{secθ+1}{secθ-1}}

\frac{secθ-1+secθ+1}{\sqrt{sec^2θ-1}}

\frac{2secθ}{tanθ}

\frac{2}{cosθ}×\frac{cosθ}{sinθ}



\frac{2}{sinθ}      

= 2 cosec θ

= R.H.S.

Hence proved.

(ii) \sqrt{\frac{1+sinθ}{1-sinθ}}+\sqrt{\frac{1-sinθ}{1+sinθ}}=2secθ

Solution:

We have,

L.H.S. = \sqrt{\frac{1+sinθ}{1-sinθ}}+\sqrt{\frac{1-sinθ}{1+sinθ}}

\frac{1+sinθ+1-sinθ}{\sqrt{1-sin^2θ}}

\frac{2}{cosθ}

= 2 sec θ

= R.H.S.

Hence proved.

(iii) \sqrt{\frac{1+cosθ}{1-cosθ}}+\sqrt{\frac{1-cosθ}{1+cosθ}}=2cosecθ      

Solution:

We have,

L.H.S. = \frac{1+cosθ+1-cosθ}{\sqrt{1-cos^2θ}}

\frac{2}{sinθ}

= 2 cosec θ

= R.H.S.

Hence proved.

(iv) \frac{secθ-1}{secθ+1}=(\frac{sinθ}{1+cosθ})^2

Solution:



We have,

L.H.S. = \frac{secθ-1}{secθ+1}

\frac{1-cosθ}{1+cosθ}

\frac{(1-cosθ)(1+cosθ)}{(1+cosθ)(1+cosθ)}

\frac{1-cos^2θ}{(1+cosθ)^2}

(\frac{sinθ}{1+cosθ})^2

= R.H.S.

Hence proved.

Question 39. (sec A – tan A)2 = \frac{1-sinA}{1+sinA}

Solution:

We have,

L.H.S. = (sec A – tan A)2

(\frac{1}{cosA}-\frac{sinA}{cosA})^2

\frac{(1-sinA)^2}{cos^2A}

\frac{(1-sinA)^2}{1-sin^2A}

\frac{(1-sinA)^2}{(1+sinA)(1-sinA)}

\frac{1-sinA}{1+sinA}

= R.H.S.

Hence proved.

Question 40. \frac{1-cosA}{1+cosA}=(cotA-cosecA)^2

Solution:

We have,



L.H.S. = \frac{1-cosA}{1+cosA}

\frac{(1-cosA)(1-cosA)}{(1+cosA)(1-cosA)}      

\frac{(1-cosA)^2}{1-cos^2A}

(\frac{1-cosA}{sinA})^2

= (cosec A – cot A)2

= (cot A – cosec A)2

= R.H.S.

Hence proved.

Question 41. \frac{1}{secA-1}+\frac{1}{secA+1}=2cosecAcotA

Solution:

We have,

L.H.S. = \frac{1}{secA-1}+\frac{1}{secA+1}

\frac{secA+1+secA-1}{(secA-1)(secA+1)}

\frac{2secA}{sec^2A-1}

\frac{2secA}{tan^2A}

\frac{2cos^2A}{cosAsin^2A}

\frac{2cosA}{sinAsinA}

= 2 cosec A cot A

= R.H.S.

Hence proved.

Question 42. \frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}=sinA+cosA

Solution:



We have,

L.H.S. = \frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}      

\frac{cosA}{1-\frac{sinA}{cosA}}+\frac{sinA}{1-\frac{cosA}{sinA}}

\frac{cos^2A}{cosA-sinA}+\frac{sin^2A}{sinA-cosA}

\frac{cos^2A}{cosA-sinA}-\frac{sin^2A}{cosA-sinA}

\frac{cos^2A-sin^2A}{cosA-sinA}

\frac{(cosA+sinA)(cosA-sinA)}{cosA-sinA}

= sin A + cos A

= R.H.S.

Hence proved.

Question 43. \frac{cosecA}{cosecA-1}+\frac{cosecA}{cosecA+1}=2sec^2A      

Solution:

We have,

L.H.S. = \frac{cosecA}{cosecA-1}+\frac{cosecA}{cosecA+1}

\frac{cosecA(cosecA+1+cosecA-1)}{cosec^2A-1}

\frac{2cosec^2A}{cot^2A}

\frac{2sin^2A}{sin^2A.cos^2A}

\frac{2}{cos^2A}

= 2 sec2 A

= R.H.S.

Hence proved.



Question 44. \frac{tan^2A}{1+tan^2A}+\frac{cot^2A}{1+cot^2A}=1

Solution:

We have,

L.H.S. = \frac{tan^2A}{1+tan^2A}+\frac{cot^2A}{1+cot^2A}      

\frac{\frac{sin^2A}{cos^2A}}{\frac{cos^2A+sin^2A}{cos^2A}}+\frac{\frac{cos^2A}{sin^2A}}{\frac{sin^2A+cos^2A}{sin^2A}}

\frac{sin^2A}{cos^2A+sin^2A}+\frac{cos^2A}{cos^2A+sin^2A}

\frac{cos^2A+sin^2A}{cos^2A+sin^2A}

= 1

= R.H.S.

Hence proved.

Question 45. \frac{cotA-cosA}{cotA+cosA}=\frac{cosecA-1}{cosecA+1}

Solution:

We have,

L.H.S. = \frac{cotA-cosA}{cotA+cosA}      

\frac{\frac{cosA}{sinA}-cosA}{\frac{cosA}{sinA}+cosA}

\frac{\frac{cosA}{sinA}-cosA}{\frac{cosA}{sinA}+cosA}

\frac{cosAcosecA-cosA}{cosAcosecA+cosA}

\frac{cosA(cosecA-1)}{cosA(cosecA+1)}

\frac{cosecA-1}{cosecA+1}

= R.H.S.

Hence proved.

Question 46. \frac{1+cosθ-sin^2θ}{sinθ(1+cosθ)}=cotθ

Solution:



We have,

L.H.S. = \frac{1+cosθ-sin^2θ}{sinθ(1+cosθ)}

\frac{(1-sin^2θ)+cosθ}{sinθ(1+cosθ)}

\frac{cos^2θ+cosθ}{sinθ(1+cosθ)}

\frac{cosθ(1+cosθ)}{sinθ(1+cosθ)}

\frac{cosθ(1+cosθ)}{sinθ(1+cosθ)}

= cos θ/sin θ

= cot θ

= R.H.S.

Hence proved.

Question 47. (i) \frac{1+cosθ+sinθ}{1+cosθ-sinθ}=\frac{1+sinθ}{cosθ}

Solution:

We have,

L.H.S. = \frac{1+cosθ+sinθ}{1+cosθ-sinθ}

\frac{\frac{1+cosθ+sinθ}{cosθ}}{\frac{1+cosθ-sinθ}{cosθ}}

\frac{secθ+1+tanθ}{secθ+1-tanθ}

\frac{\frac{1}{secθ-tanθ}+1}{secθ-tanθ+1}

\frac{1+secθ-tanθ}{1+secθ-tanθ}×\frac{1}{secθ-tanθ}

\frac{1}{secθ-tanθ}

\frac{secθ+tanθ}{(secθ-tanθ)(secθ+tanθ)}

= sec θ + tan θ



= 1/cos θ + sin θ/cos θ

\frac{1+sinθ}{cosθ}

= R.H.S.

Hence proved.

(ii) \frac{sinθ-cosθ+1}{sinθ+cosθ-1}=\frac{1}{secθ-tanθ}

Solution:

We have,

L.H.S. = \frac{sinθ-cosθ+1}{sinθ+cosθ-1}

\frac{\frac{sinθ-cosθ+1}{cosθ}}{\frac{sinθ+cosθ-1}{cosθ}}

\frac{tanθ-1+secθ}{tanθ+1-secθ}

\frac{(tanθ+secθ)-1}{tanθ+1-secθ}

\frac{\frac{1}{secθ-tanθ}-1}{tanθ+1-secθ}

\frac{\frac{1-secθ+tanθ}{secθ-tanθ}}{tanθ+1-secθ}

\frac{1-secθ+tanθ}{1-secθ+tanθ}×\frac{1}{secθ-tanθ}

\frac{1}{secθ-tanθ}

= R.H.S.

Hence proved.

Question 48. \frac{cosθ-sinθ+1}{cosθ+sinθ-1}=cosecθ+cotθ

Solution:

We have,

L.H.S. = \frac{cosθ-sinθ+1}{cosθ+sinθ-1}

\frac{\frac{cosθ-sinθ+1}{sinθ}}{\frac{cosθ+sinθ-1}{sinθ}}

\frac{cotθ-1+cosecθ}{cotθ+1-cosecθ}

\frac{(cotθ+cosecθ)-1}{cotθ-cosecθ+1}

\frac{\frac{1}{cosecθ-cotθ}-1}{cotθ-cosecθ+1}

\frac{\frac{1-cosecθ+cotθ}{cosecθ-cotθ}}{1-cosecθ+cotθ}

\frac{1-cosecθ+cotθ}{1-cosecθ+cotθ}×\frac{1}{cosecθ-cotθ}

\frac{1}{cosecθ-cotθ}

\frac{cosecθ+cotθ}{(cosecθ-cotθ)(cosecθ+cotθ)}

= cosec θ + cot θ

= R.H.S.

Hence proved.



Question 49. (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ

Solution:

We have,

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

= sin2 θ/cosθ + cos θ + sin θ + cos2 θ/sin θ 

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

= (1 + tan θ) (sin θ + cos θ/tan θ)

\frac{sin^2θ+cos^2θ}{sinθ}(tanθ+1)

\frac{tanθ+1}{sinθ}

= sec θ + cosec θ

= R.H.S.

Hence proved.

Question 50. \frac{tanA}{1+secA}-\frac{tanA}{1-secA}=2cosecA

Solution:

We have,

L.H.S. = \frac{tanA}{1+secA}-\frac{tanA}{1-secA}

\frac{\frac{sinA}{cosA}}{1+\frac{1}{cosA}}-\frac{\frac{sinA}{cosA}}{1-\frac{1}{cosA}}

\frac{\frac{sinA}{cosA}}{\frac{cosA+1}{cosA}}-\frac{\frac{sinA}{cosA}}{\frac{cosA-1}{cosA}}

\frac{sinA}{cosA+1}-\frac{sinA}{cosA-1}

sinA(\frac{1}{cosA+1}-\frac{1}{cosA-1})

sinA(\frac{cosA-1-cosA-1}{cos^2A-1})

sinA(\frac{-2}{-sin^2A})

= 2/sin A

= 2 cosec A

= R.H.S.

Hence proved.

Question 51. 1 + \frac{cot^2θ}{1+cosecθ}  = cosec θ

Solution:

We have,

L.H.S. = 1 + \frac{cot^2θ}{1+cosecθ}

= 1 + \frac{cosec^2θ-1}{1+cosecθ}

= 1+ \frac{(cosecθ+1)(cosecθ-1)}{1+cosecθ}

= 1 + cosec θ − 1



= cosec θ

= R.H.S.

Hence proved.

Question 52. \frac{cosθ}{cosecθ+1}+\frac{cosθ}{cosecθ-1}=2tanθ

Solution:

We have,

L.H.S. = \frac{cosθ}{cosecθ+1}+\frac{cosθ}{cosecθ-1}

\frac{cosθ}{\frac{1}{sinθ}+1}+\frac{cosθ}{\frac{1}{sinθ}-1}

\frac{cosθsinθ}{1+sinθ}+\frac{cosθsinθ}{1-sinθ}

cosθsinθ(\frac{1}{1+sinθ}+\frac{1}{1-sinθ})

cosθsinθ(\frac{2}{1-sin^2θ})

cosθsinθ(\frac{2}{cos^2θ})

= 2 sin θ/cos θ

= 2 tan θ

= R.H.S.

Hence proved.

Question 53. (1 + tan2 A) + (1 + 1/tan2 A) = 1/(sin2 A − sin4 A)

Solution:

We have,

L.H.S. = (1 + tan2 A) + (1 + 1/tan2 A)

= (1 + sin2 A/cos2 A) + (1 + cos2 A/sin2 A)

= 1/cos2 A + 1/sin2 A

\frac{cos^2A+sin^2A}{cos^2Asin^2A}

\frac{1}{sin^2A(1-sin^2A)}

= 1/(sin2 A − sin4 A)

= R.H.S.

Hence proved.

Question 54. sin2 A cos2 B cos2 A sin2 B = sin2 A sin2 B  

Solution:

We have,

L.H.S. = sin2 A cos2 B − cos2 A sin2 B

= sin2 A (1 − sin2 B) − sin2 B (1 − sin2 A)

= sin2 A− sin2 A sin2 B − sin2 B + sin2 A sin2 B  

= sin2A − sin2 B

= R.H.S. 

Hence Proved. 

Question 55. (i) \frac{cotA+tanB}{cotB+tanA}=cotAtanB     

Solution:

We have,

L.H.S. = \frac{cotA+tanB}{cotB+tanA}

\frac{\frac{cosA}{sinA}+\frac{sinB}{cosB}}{\frac{cosB}{sinB}+\frac{sinA}{cosA}}

\frac{cosAcosB+sinAsinB}{sinAcosB}×\frac{cosAsinB}{cosAcosB+sinAsinB}

\frac{cosAsinB}{sinAcosB}

= cot A tan B



= R.H.S.

Hence proved.

(ii) \frac{tanA+tanB}{cotA+cotB}=tanAtanB

Solution:

We have,

L.H.S. = \frac{tanA+tanB}{cotA+cotB}

\frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{\frac{cosA}{sinA}+\frac{cosB}{sinB}}

\frac{sinAcosB+sinBcosA}{cosAcosB}×\frac{sinAsinB}{sinAcosB+sinBcosA}

\frac{sinAsinB}{cosAcosB}

= tan A tan B

= R.H.S.

Hence proved. 

Question 56. cot2 A cosec2 B − cot2 B cosec2 A = cot2 A − cot2B

Solution:

We have,

L.H.S. = cot2 A cosec2 B − cot2 B cosec2 A

= cot2 A (1 + cot2 B) − cot2 B (1 + cot2 A)

= cot2 A + cot2 A cot2 B − cot2 B − cot2 B cot2 A  

= cot2 A − cot2 B

= R.H.S.

Hence proved.

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