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

Chapter 5 of RD Sharma’s Class 11 Mathematics textbook covers Trigonometric Functions in which form the foundation for understanding trigonometry. This chapter explores the various trigonometric functions and their properties providing the crucial base for higher-level mathematics and applications in science and engineering.

Trigonometric Functions

The Trigonometric functions relate the angles of a triangle to the lengths of its sides. They are fundamental in the various fields of mathematics and applied sciences. The primary trigonometric functions are sine, cosine, tangent, cosecant, secant and cotangent. These functions are defined based on the angles and sides of the right-angled triangles and can be extended to the all real numbers through the unit circle approach.

Prove the following identities (1 - 13)

Question 1. sec⁴θ - sec²θ = tan⁴θ + tan²θ  

Solution:

We have

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)

Read More:

• Six Trigonometric Functions
• Trigonometric Functions