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Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.1 | Set 2
  • Last Updated : 21 Feb, 2021

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}

= cos6/3θ + sin6/3θ

= 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 Tn = 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 Tn = 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 Tn = 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

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