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Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.1 | Set 1

  • Last Updated : 21 Feb, 2021

Prove the following identities (1 – 13)

Question 1. sec4θ – sec2θ = tan4θ + tan2θ  

Solution:

We have

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sec4θ – sec2θ = tan4θ + tan2θ



Taking LHS

= sec4θ – sec2θ

= sec2θ(sec2θ – 1)

Using sec2 θ = tan2θ + 1, we get 

= (1 + tan2θ)tan2θ         

= tan2θ + tan4θ

Hence, LHS = RHS (Proved)

Question 2. sin6θ + cos6θ = 1 – 3sin2θcos2θ      

Solution:



We have

sin6θ + cos6θ = 1 – 3sin2θcos2θ 

Taking LHS

= sin6θ + cos6θ   

= (sin2θ)3 + (cos2θ)3

Using a3 + b3 = (a + b)(a2 + b2 – ab), we get

= (sin2θ + cos2θ)(sin4θ + cos4θ – sin2θcos2θ)     

Using a2 + b2 = (a + b)2 – 2ab and sin2θ + cos2θ = 1, we get

= (1)[(sin2θ + cos2θ)2 – 2sin2θcos2θ – sin2θcos2θ]             

= (1)[(1)2 – 3sin2θcos2θ] 



= 1 – 3sin2θcos2θ

Hence, LHS = RHS (Proved)

Question 3. (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

Solution:

We have

(cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

Taking LHS

= (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ)

Using cosecθ = 1/sinθ and secθ = 1/cosθ   

(\frac{1}{sinθ} - sin θ)(\frac{1}{cosθ}-cos θ)(\frac{sinθ}{cosθ}+\frac{cosθ}{sinθ})          

(\frac{1- sin^2 θ}{sinθ} )(\frac{1- cos^2 θ}{cosθ})(\frac{sin^2θ+cos^2 θ}{cosθsinθ})         



(\frac{cos^2 θ}{sinθ} )(\frac{sin^2θ}{cosθ})(\frac{1}{cosθsinθ})

= 1

Hence, LHS = RHS (Proved)

Question 4. cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

Solution:

We have

cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

Taking LHS

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

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

\frac{(1-cosθ)}{sinθ}[(\frac{1}{cosθ})-cosθ]



\frac{(1-cosθ)}{sinθ}(\frac{1-cos^2θ}{cosθ})

\frac{(1-cosθ)}{sinθ}(\frac{sin^2θ}{cosθ})

\frac{sinθ}{cosθ}-sinθ

\frac{sinθ}{cosθ}-sinθ

Hence, LHS = RHS(Proved)

Question 5. \frac{1-sinAcosA}{cosA(secA-cosecA)}.\frac{sin ^2A-cos^2A}{sin^3A+cos^3A}=sinA

Solution:

We have

\frac{1-sinAcosA}{cosA(secA-cosecA)}.\frac{sin ^2A-cos^2A}{sin^3A+cos^3A}=sinA

Taking LHS

\frac{1-sinAcosA}{cosA(\frac{1}{cosA}-\frac{1}{sinA})}.\frac{(sin A)^2-(cosA)^2}{(sinA)^3+(cosA)^3}              



Using a2 – b2 = (a + b)(a – b) and a3 + b3 = (a + b)(a2 + b2ab), we get

\frac{1-sinAcosA}{cosA(\frac{sinA-cosA}{cosAsinA})}.\frac{(sin A+cosA)(sinA-cosA)}{(sinA+cosA)(sin^2A+cos^2A-sinAcosA)}

\frac{sinA(1-sinAcosA)}{(sinA-cosA)}.\frac{(sin A+cosA)(sinA-cosA)}{(sinA+cosA)(sin^2A+cos^2A-sinAcosA)}

\frac{sinA(1-sinAcosA)}{1}.\frac{1}{(sin^2A+cos^2A-sinAcosA)}

\frac{sinA(1-sinAcosA)}{1}.\frac{1}{(1-sinAcosA)}

= sinA

Hence, LHS = RHS(Proved)

Question 6. \frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=(secAcosecA+1)

Solution:

We have

\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=(secAcosecA+1)



Taking LHS

\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}

Using tanA = sinA/cosA and cotA = cosA/sinA, we get

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

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

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

\frac{sin^3A-cos^3A}{sinAcosA(sinA-cosA)}

Using a3 – b3 = (a – b)(a2 + b2 + ab), we get

\frac{(sinA-cosA)[(sinA)^2+(cosA)^2+sinAcosA]}{sinAcosA(sinA-cosA)}             

\frac{(1+sinAcosA)}{sinAcosA}



\frac{1}{sinAcosA}+\frac{sinAcosA}{sinAcosA}

Using cosecA = 1/sinA and secA = 1/cosA, we get

= secAcosecA + 1         

Hence, LHS = RHS(Proved)                

Question 7. \frac{sin^3A+cos^3A}{sinA+cosA}+\frac{sin^3A-cos^3A}{sinA-cosA}=2

Solution:  

We have

\frac{sin^3A+cos^3A}{sinA+cosA}+\frac{sin^3A-cos^3A}{sinA-cosA}=2

Taking LHS

\frac{sin^3A+cos^3A}{sinA+cosA}+\frac{sin^3A-cos^3A}{sinA-cosA}

Using a3 ± b3 = (a ± b)(a2 + b2 ± ab), we get

\frac{(sinA+cosA)((sinA)^2+(cosA)^2-sinAcosA)}{sinA+cosA}+\frac{(sinA-cosA)((sinA)^2+(cosA)^2+sinAcosA)}{sinA-cosA}         

Using sin2θ + cos2θ = 1, we get

= 1 – sinAcosA + 1 + sinAcosA          

= 2

Hence, LHS = RHS(Proved)

Question 8. (secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2 = 1

Solution:

We have

(secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2 = 1

Taking LHS

= (secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2



Expanding the above equation using the formula  

(a + b)2 = a2 + b2 + 2ab

= (secAsecB)2 + (tanAtanB)2 + 2(secAsecB)(tanAtanB) – 

   (secAtanB)2 – (tanAsecB)2 – 2(secAtanB)(tanAsecB)

= sec2Asec2B + tan2Atan2B – sec2Atan2B – tan2Asec2B

= sec2A(sec2B – tan2B) – tan2A(sec2B – tan2B)

= sec2A – tan2A                -(Using sec2θ – tan2θ = 1)

= 1

Hence, LHS = RHS(Proved)

Question 9. \frac{cosθ}{1-sinθ}=\frac{1+cosθ+sinθ}{1+cosθ-sinθ}

Solution:

We have

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

Taking RHS

\frac{1+cosθ+sinθ}{1+cosθ-sinθ}

\frac{(1+cosθ)+sinθ}{(1+cosθ)-sinθ}

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

\frac{((1+cosθ)+sinθ)^2}{(1+cosθ)^2-sin^2θ}

\frac{(1+cosθ)^2+sin^2θ+2(1+cosθ)(sinθ)}{(1)^2+(cosθ)^2+2cosθ-(sin^2θ)}

\frac{(1)^2+(cosθ)^2+2cosθ+sin^2θ+2sinθ+2cosθsinθ}{(1)^2+(cosθ)^2+2cosθ-(sin^2θ)}

\frac{1+cos^2θ+sin^2θ+2cosθ+2sinθ+2cosθsinθ}{(1-sin^2θ)+(cosθ)^2+2cosθ}



\frac{1+1+2cosθ+2sinθ+2cosθsinθ}{cos^2θ+cos^2θ+2cosθ}

\frac{2(1+cosθ+sinθ+cosθsinθ)}{2cosθ(cosθ+1)}

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

\frac{(1+cosθ)(1+sinθ)}{cosθ(cosθ+1)}

\frac{(1+sinθ)}{cosθ}

\frac{(1+sinθ)}{cosθ} ×\frac{cosθ}{cosθ}

\frac{(1+sinθ)(cosθ)}{cos^2θ}

\frac{(1+sinθ)(cosθ)}{1-sin^2θ}

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

\frac{cosθ}{(1-sinθ)}



Hence, RHS = LHS(Proved)

Question 10. \frac{tan^3x}{1+tan^2x} + \frac{cot^3x}{1+cot^2x}=\frac{1-2sin^2xcos^2x}{sinxcosx}

Solution:

We have

\frac{tan^3x}{1+tan^2x} + \frac{cot^3x}{1+cot^2x}=\frac{1-2sin^2xcos^2x}{sinxcosx}

Taking LHS

\frac{tan^3x}{1+tan^2x} + \frac{cot^3x}{1+cot^2x}

Using 1 + tan2x = sec2x and 1 + cot2x = cosec2x, we get

\frac{tan^3x}{sec^2x} + \frac{cot^3x}{cosec^2x}

\frac{\frac{sin^3x}{cos^3x}}{\frac{1}{cos^2x}} + \frac{\frac{cos^3x}{sin^3x}}{\frac{1}{sin^2x}}

\frac{sin^3x}{cosx} + \frac{cos^3x}{sinx}

\frac{sin^4x + cos^4x}{sinxcosx}

\frac{(sin^2x)^2 + (cos^2x)^2}{sinxcosx}

Using a2 + b2 = (a + b)2 – 2ab, we get

\frac{(sin^2x+cos^2x)^2 - 2sin^2xcos^2x}{sinxcosx}

\frac{(1)^2-2sin^2θcos^2θ}{sinθcosθ}

\frac{1-2sin^2xcos^2x}{sinxcosx}

Hence, LHS = RHS (Proved)

Question 11. 1-\frac{sin^2θ}{1+cotθ}-\frac{cos^2θ}{1+tanθ}=sinθcosθ

Solution:

We have

1-\frac{sin^2θ}{1+cotθ}-\frac{cos^2θ}{1+tanθ}=sinθcosθ



Taking LHS

1-\frac{sin^2θ}{1+cotθ}-\frac{cos^2θ}{1+tanθ}

By using the formulas cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, we get

1-\frac{sin^2θ}{1+\frac{cosθ}{sinθ}}-\frac{cos^2θ}{1+\frac{sinθ}{cosθ}}

1-\frac{sin^3θ}{sinθ+cosθ}-\frac{cos^3θ}{cosθ+sinθ}

\frac{cosθ+sinθ-sin^3θ-cos^3θ}{sinθ+cosθ}

Using a3+b3 = (a + b)(a2 + b2 – ab), we get  

\frac{cosθ+sinθ-(sinθ+cosθ)(sin^2θ+cos^2θ-sinθcosθ)}{sinθ+cosθ}                                         

\frac{(cosθ+sinθ)(1-sin^2θ-cos^2θ+sinθcosθ)}{(sinθ+cosθ)}

= 1 – (sin2θ + cos2θ) + sinθcosθ

= 1 – 1 + sinθcosθ

= sinθcosθ

Hence, LHS = RHS (Proved)

Question 12.(\frac{1}{sec^2θ-cos^2θ}+\frac{1}{cosec^2θ-sin^2θ})sin^2θcos^2θ=\frac{1-sin^2θcos^2θ}{2+sin^2θcos^2θ}

Solution:

We have

(\frac{1}{sec^2θ-cos^2θ}+\frac{1}{cosec^2θ-sin^2θ})sin^2θcos^2θ=\frac{1-sin^2θcos^2θ}{2+sin^2θcos^2θ}

Taking LHS

(\frac{1}{sec^2θ-cos^2θ}+\frac{1}{cosec^2θ-sin^2θ})sin^2θcos^2θ

(\frac{1}{\frac{1}{cos^2θ}-cos^2θ}+\frac{1}{\frac{1}{sin^2θ}-sin^2θ})sin^2θcos^2θ            

(\frac{cos^2θ}{1-cos^4θ}+\frac{sin^2θ}{1-sin^4θ})sin^2θcos^2θ



(\frac{cos^2θ(1-sin^4θ)+sin^2θ(1-cos^4θ)}{(1-cos^4θ)(1-sin^4θ)})sin^2θcos^2θ

(\frac{cos^2θ-cos^2θsin^4θ+sin^2θ-sin^2θcos^4θ)}{(1-cos^2θ)(1+cos^2θ)(1-sin^2θ)(1+sin^2θ)})sin^2θcos^2θ

(\frac{1-cos^2θsin^4θ-sin^2θcos^4θ)}{sin^2θ(1+cos^2θ)cos^2θ(1+sin^2θ)})sin^2θcos^2θ

(\frac{1-cos^2θsin^2θ(sin^2θ+cos^2θ)}{(1+cos^2θ)(1+sin^2θ)})

\frac{1-cos^2θsin^2θ}{(1+cos^2θ+sin^2θ+cos^2θsin^2θ)}

\frac{1-cos^2θsin^2θ}{2+cos^2θsin^2θ}

Hence, LHS = RHS(Proved)

Question 13. (1 + tanαtanβ)2 + (tanα – tanβ)2  = sec2αsec2β

Solution:

We have

(1 + tanαtanβ)2 + (tanα – tanβ)2 = sec^2αsec2β

Taking LHS

= (1 + tanαtanβ)2 + (tanα – tanβ)2

= (1 + tan2αtan2β + 2tanαtanβ) + (tan2α + tan2β – 2tanαtanβ)

= 1 + tan2αtan2β + tan2α + tan2β

= (1 + tan2β) + tan2α(1 + tan2β)

= (1 + tan2β)(1 + tan2α)

= sec2αsec2β

Hence, LHS = RHS (Proved)




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