Class 11 RD Sharma Solutions - Chapter 5 Trigonometric Functions - Exercise 5.1 | Set 2

Question 14. Prove that \frac{(1+cotθ+tanθ)(sinθ-cosθ)}{sec^3θ-cosec^3θ}=sin^2θcos^2θ  

Solution:

We have

\frac{(1+cotθ+tanθ)(sinθ-cosθ)}{sec^3θ-cosec^3θ}=sin^2θcos^2θ  

Taking LHS

= \frac{(1+cotθ+tanθ)(sinθ-cosθ)}{sec^3θ-cosec^3θ}

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

= \frac{(\frac{cosθsinθ+cos^2θ+sin^2θ}{sinθcosθ})(sinθ-cosθ)}{\frac{sin^3θ-cos^3θ}{sin^3θcos^3θ}}

= \frac{(1+cosθsinθ)(sinθ-cosθ)(sin^2θcos^2θ)}{sin^3θ-cos^3θ}

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

= \frac{(1+cosθsinθ)(sin^2θcos^2θ)}{(1+cosθsinθ)}

= sin²θcos²θ

Hence, LHS = RHS (Proved)

Question 15. Prove that \frac{2sinθcosθ-cosθ}{1-sinθ+sin^2θ-cos^2θ}=cotθ

Solution:

We Have

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

Taking LHS

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

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

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

= \frac{cosθ(2sinθ-1)}{2sin^2θ-sinθ}

= \frac{cosθ(2sinθ-1)}{sinθ(2sinθ-1)}

= cosθ/sinθ

= cotθ

Hence, LHS = RHS(Proved)

Question 16. Prove that cosθ(tanθ + 2)(2tanθ + 1) = 2secθ + 5sinθ

Solution:

We have

cosθ(tanθ + 2)(2tanθ + 1) = 2secθ + 5sinθ

Taking LHS

= cosθ(tanθ + 2)(2tanθ + 1)

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

= cosθ\frac{(sinθ+2cosθ)(2sinθ+cosθ)}{cos^2θ}

= \frac{(2sin^2θ+sinθcosθ+4sinθcosθ+2cos^2θ)}{cosθ}

= \frac{2(sin^2θ+cos^2θ)+5sinθcosθ}{cosθ}

= \frac{2+5sinθcosθ}{cosθ}

= \frac{2}{cosθ}+\frac{5sinθcosθ}{cosθ}

= 2secθ + 5sinθ

Hence, LHS = RHS(Proved)

Question 17. If x = \frac{2sinθ}{1+cosθ+sinθ} , prove that \frac{1-cosθ+sinθ}{1+sinθ}  is also equal to x.

Solution:

We have 

 x = \frac{2sinθ}{1+cosθ+sinθ}

Taking LHS

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

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

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

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

=  \frac{1+cosθ-sinθ}{1+sinθ}

Question 18. If sin θ=\frac{a^2+b^2}{a^2−b^2} , then find the values of tanθ, secθ, and cosecθ

Solution: 

We have 

sin θ=\frac{Perpendicular}{Hypotenuse}=\frac{a^2+b^2}{a^2−b^2}  

As we know that 

cosθ = √1 - sin²θ         -(1)

Now put the value of sinθ in eq(1)

cosθ = \sqrt{1-\frac{(a^2-b^2)^2}{(a^2+b^2)^2}}

= \sqrt{\frac{(a^2+b^2)^2-(a^2-b^2)^2}{(a^2+b^2)^2}}

= \sqrt{\frac{(a^4+b^4+2a^2b^2)-(a^4+b^4-2a^2b^2)}{(a^2+b^2)^2}}

= \sqrt{\frac{4a^2b^2}{(a^2+b^2)^2}}

= \frac{2ab}{(a^2+b^2)}

So the value of cosθ = \frac{2ab}{(a^2+b^2)}

Now,

tanθ = \frac{Perpendicular}{Base}=\frac{a^2−b^2}{2ab}

secθ = \frac{Hypotenuse}{Base}=\frac{a^2+b^2}{2ab}

cosecθ = \frac{Hypotenuse}{Perpendicular}=\frac{a^2+b^2}{a^2-b^2}

Alternative Method:

We have 

sin θ=\frac{Perpendicular}{Hypotenuse}=\frac{a^2+b^2}{a^2−b^2}

We draw a △PQR right-angled at Q PR = a² + b² and PQ = a² - b²

By Pythagoras theorem, we have

PR² = PQ² + QR²

QR² = (a² + b²)² - (a² - b²)²

QR² = (a⁴ + b⁴ + 2a²b²) − (a⁴ + b⁴ − 2a²b²)

QR² = 4a²b²

QR = 2ab

cosθ = \frac{2ab}{a^2 +b^2}

Now,

tanθ = \frac{Perpendicular}{Base}=\frac{a^2−b^2}{2ab}

secθ = \frac{Hypotenuse}{Base}=\frac{a^2+b^2}{2ab}

cosecθ = \frac{Hypotenuse}{Perpendicular}=\frac{a^2+b^2}{a^2-b^2}

Question 19.  If tanθ = a/b, then find the value of \sqrt{\frac{a+b}{a-b}}+ \sqrt{\frac{a-b}{a+b}}

Solution: 

We have

= \sqrt{\frac{a+b}{a-b}}+ \sqrt{\frac{a-b}{a+b}}

=\sqrt{\frac{\frac{a}{b}+1}{\frac{a}{b}-1}}+ \sqrt{\frac{\frac{a}{b}-1}{\frac{a}{b}+1}}

Now put tanθ = a/b

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

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

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

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

= \frac{2sinθ}{\sqrt{sin^2θ-cos^2θ}}

Question 20. If tanθ = a/b, show that  \frac{asinθ-bcosθ}{asinθ+bcosθ}=\frac{a^2-b^2}{a^2+b^2} .

Solution:

We have

 \frac{asinθ-bcosθ}{asinθ+bcosθ}=\frac{a^2-b^2}{a^2+b^2}

Taking LHS

= \frac{asinθ-bcosθ}{asinθ+bcosθ}

Dividing denominator and Numerator by cosθ

= \frac{\frac{asinθ-bcosθ}{cosθ}}{\frac{asinθ+bcosθ}{cosθ}}

= \frac{\frac{asinθ}{cosθ}-\frac{bcosθ}{cosθ}}{\frac{asinθ}{cosθ}+\frac{bcosθ}{cosθ}}

= \frac{atanθ-b}{atanθ+b}

= \frac{a(\frac{a}{b})-b}{a(\frac{a}{b})+b}

= \frac{\frac{a^2-b^2}{b}}{\frac{a^2+b^2}{b}}

=\frac{a^2-b^2}{a^2+b^2}

Hence, LHS = RHS(Proved)

Question 21. If cosecθ - sinθ = a³, secθ - cosθ = b³, then prove that a²b²(a² + b²) = 1.

Solution: 

Given: cosecθ - sinθ = a³

1/sinθ − sinθ = a³

\frac{1-sin^2θ}{sinθ}  = a³   

cos²θ/sinθ = a³

a = (cos²θ/sinθ)^1/3 

Similarly, b = (sin²θ/cosθ)^1/3  

Now putting the values of a and b in the following equation

Taking LHS

= a²b²(a² + b²)

= a⁴b² + a²b⁴

= (\frac{cos^2θ}{sinθ})^\frac{4}{3}(\frac{sin^2θ}{cosθ})^\frac{2}{3}+ (\frac{cos^2θ}{sinθ})^\frac{2}{3}(\frac{sin^2θ}{cosθ})^\frac{4}{3}

= cos^6/3θ + sin^6/3θ

= cos²θ + sin²θ

= 1

Hence, LHS = RHS (Proved)

Question 22. If cotθ(1 + sinθ) = 4m and cotθ(1 − sinθ) = 4n, prove that (m² - n²)² = mn.

Solution: 

Given: cotθ(1 + sinθ) = 4m and cotθ(1 − sinθ) = 4n

Multiplying both the equations

16mn = cot²θ(1 - sin²θ)

16mn = \frac{cos^2θ}{sin^2θ}(cos^2θ)

16mn = cos⁴θ/sin²θ

mn = cos⁴θ/16sin²θ          -(1)

Now squaring the given equations

16m² = cot²θ(1 + sinθ)² and 16n² = cot²θ(1 - sinθ)²

On subtracting both the equation, we get

16m² - 16n² = cot²θ(1 + sinθ)² - cot²θ(1 - sinθ)²

16(m² - n²) = cot²θ((1 + sinθ)² - (1 - sinθ)²)

16(m² - n²) = \frac{cos^2θ}{sin^2θ}(4sinθ)

(m² - n²) = cos²θ/4sinθ  

On squaring both side, we get

(m² - n²)² = cos⁴θ/16sinθ          -(2)

From equation(1) and (2)

(m² - n²)² = mn   

Hence proved

Question 23. If sinθ + cosθ = m then prove that sin⁶θ + cos⁶θ = \frac{4−3(m^2−1)^2}{4} , where m² ≤ 2.

Solution:

Given: sinθ + cosθ = m

On squaring both side, we get

(sinθ + cosθ)² = m²

= sin²θ + cos²θ + 2sinθcosθ = m²              

= 2sinθcosθ = m² − 1                                       

Now,

Taking LHS

= sin⁶θ + cos⁶θ

Using a³ + b³ = (a + b)(a² + b² − ab)

= (sin²θ)³ + (cos²θ)³                   

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

= (1)((sin²θ)² + (cos²θ)² − sin²θcos²θ)

= (sin²θ + cos²θ)² − 2sin²θcos²θ − sin²θcos²θ

= (1 − 3sin²θcos²θ)

= 1−3(\frac{m^2-1}{2})^2

= 1−3\frac{(m^2-1)^2}{4}

= \frac{4-3(m^2-1)^2}{4}

Hence, Proved.

Question 24. If a = secθ - tanθ and b = cosecθ + cotθ, then show that ab + a - b + 1 = 0.

Solution:

We have

 a = secθ - tanθ and b = cosecθ + cotθ

and we have to proof that 

 ab + a - b + 1 = 0

So, taking LHS

 ab + a - b + 1 

Now put the values of a and b, we get

= (secθ - tanθ)(cosecθ + cotθ) - (secθ - tanθ) + (cosecθ + cotθ) + 1

= (1/cosθ - sinθ/cosθ)(1/sinθ + cosθ/sinθ) - (1/cosθ - sinθ/cosθ) + (1/sinθ + cosθ/sinθ) + 1

= 1/cosθsinθ + 1/cosθ x cosθ/sinθ - sinθ/cosθ x 1/sinθ - (sinθ/cosθ) x (cosθ/sinθ) + 1/cosθ - sinθ/cosθ - 1/sinθ - cosθ/sinθ + 1

= 1/cosθsinθ + 1/sinθ - 1/cosθ - 1 + 1/cosθ - sinθ/cosθ - 1/sinθ - cosθ/sinθ + 1

= 1/cosθsinθ - sinθ/cosθ - cosθ/sinθ 

= 1 - sin²θ - cos²θ/sinθcosθ

= 1 - (sin²θ + cos²θ)/sinθcosθ

= 1 - 1/sinθcosθ

= 0

Hence, LHS = RHS (Proved)

Question 25. |\sqrt\frac{1-sinθ}{1+sinθ}+\sqrt\frac{1+sinθ}{1-sinθ}|=\frac{-2}{cosθ} , where π/2 < θ < π. 

Solution:

We have

|\sqrt\frac{1-sinθ}{1+sinθ}+\sqrt\frac{1+sinθ}{1-sinθ}|=\frac{-2}{cosθ}

Taking LHS

= |\sqrt\frac{1-sinθ}{1+sinθ}+\sqrt\frac{1+sinθ}{1-sinθ}|

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

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

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

= \sqrt\frac{(1-sinθ)^2}{cos^2θ}+\sqrt\frac{(1+sinθ)^2}{cos^2θ}

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

= \frac{(1-sinθ+1+sinθ)}{cosθ}

= 2/cosθ 

Since π/2 < θ < π   ,where cosθ is negative

So, -2/cosθ 

Hence, LHS = RHS (Proved)

Question 26 (i). If T_n = sinⁿθ + cosⁿθ, prove that

\frac{T_3-T_5}{T_1}=\frac{T_5-T_7}{T_5}        

Solution: 

LHS = \frac{T_3-T_5}{T_1}=\frac{(sin^3θ+cos^3θ)-(sin^5θ+cos^5θ)}{sinθ+cosθ}

=\frac{sin^3θ-sin^5θ+cos^3θ-cos^5θ}{sinθ+cosθ}

=\frac{sin^3θ(1-sin^2θ)+cos^3θ(1-cos^2θ)}{sinθ+cosθ}

=\frac{sin^3θcos^2θ+cos^3θsin^2θ}{sinθ+cosθ}

=\frac{sin^2θcos^2θ(sinθ+cosθ)}{sinθ+cosθ}

= sin²θcos²θ

RHS = \frac{T_5-T_7}{T_5}     

=\frac{(sin^5θ+cos^5θ)-(sin^7θ+cos^7θ)}{sin^3θ+cos^3θ}

=\frac{sin^5θ-sin^7θ+cos^5θ-cos^7θ}{sin^3θ+cos^3θ}

=\frac{sin^5θ(1-sin^2θ)+cos^5θ(1-cos^2θ)}{sin^3θ+cos^3θ}

=\frac{sin^5θcos^2θ+cos^5θsin^2θ}{sin^3θ+cos^3θ}

=\frac{sin^2θcos^2θ(sin^3θ+cos^3θ)}{sin^3θ+cos^3θ}

= sin2θcos²θ

Question 26 (ii). If T_n = sinⁿθ + cosⁿθ, prove that

2T₆ - 3T₄ + 1 = 0

Solution: 

LHS = 2(sin⁶θ + cos⁶θ) - 3(sin⁴θ + cos⁴θ) + 1

Using (a³ + b³) = (a + b)(a² + b² - ab)

= 2(sin²θ + cos²θ)(sin⁴θ + cos⁴θ - sin²θcos²θ) - 3(sin⁴θ + cos⁴θ) + 1                             

= 2(1)(sin⁴θ + cos⁴θ - sin²θcos²θ) - 3(sin⁴θ + cos⁴θ) + 1

= 2sin⁴θ + 2cos⁴θ - 2sin²θcos²θ - 3sin⁴θ - 3cos⁴θ + 1

= -sin⁴θ - cos⁴θ - 2sin²θcos²θ + 1

= -(sin²θ + cos²θ)² + 1

= -1 + 1 = 0 = RHS (Hence Proved)

Question 26 (iii). If T_n = sinⁿθ + cosⁿθ, prove that

6T₁₀ - 15T₈ + 10T₆ - 1 = 0 

Solution:

T₆ = sin⁶θ + cos⁶θ

Using a³ + b³ = (a + b)(a² + b² − ab)

= (sin²x)³ + (cos²x)³                                                                

= (sin²x + cos²x)(sin⁴x + cos⁴x − sin²xcos²x)

Using a² + b² = (a + b)² − 2ab

= (1)(sin⁴x + cos⁴x − sin²xcos²x)                                      

= (sin²x)² + (cos²x)² − sin²xcos²x

= (sin²x + cos²x)² − 3sin²xcos²

= 1 − 3sin²xcos²x

Similarly, we get the values of T₈ & T₁₀

T₈ = (sin⁶x + cos⁶x)(sin²x + cos²x) − sin²xcos²x(sin⁴x + cos⁴x)

= 1 − 3sin²xcos²x − sin²xcos²x(1 − 2sin²xcos²x)

= 1 − 4sin²xcos²x + 2sin⁴xcos⁴x

T₁₀ = sin¹⁰θ + cos¹⁰θ

= (sin⁶θ + cos6θ)(sin⁴θ + cos⁴θ) − sin4θcos⁴θ(sin²θ + cos²θ)

= (1 − 3sin²xcos²x)(1 − 2sin²xcos²x) − sin⁴xcos⁴x

= 1 − 5sin²xcos²x + 5sin⁴xcos⁴x

On putting the values of T6, T8 and T10 in the following equation 

6T₁₀ - 15T₈ + 10T₆ - 1

We get the value 0.

Hence Proved