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Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.1 | Set 2
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Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 1

Last Updated : 09 Mar, 2022
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Prove the following identities: 

Question 1. √[(1 – cos2x)/(1 + cos2x)] = tanx

Solution:

Let us solve LHS,

= √[(1 – cos2x)/(1 + cos2x)]

As we know that, cos2x =1 – 2 sin2x

= 2 cos2x – 1

So,

= √[(1 – cos2x)/(1 + cos2x)] 

= √[(1 – (1 – 2sin2x))/(1 + (2cos2x – 1))]

= √(1 – 1 + 2sin2x)/(1 + 2cos2x – 1)1

= √[2 sin2x/2 cos2x]

= sinx / cosx

= tanx 

LHS = RHS

Hence proved.

Question 2. sin2x/(1 – cos2x) = cotx

Solution:

Let us solve LHS,

= sin 2x/(1 – cos 2x)

As we know that,

cos 2x = 1 – 2 sin2x 

Sin 2x = 2 sin x cos x

So,

sin 2x/(1-cos 2x) = (2 sin x cos x)/(1 – (1 – 2sin2x))

= (2 sin x cos x)/(1 – 1 + 2sin2x)]

= [2 sin x cos x/2 sin2x]

= cos x/sin x

= cot x

LHS = RHS

Hence proved.

Question 3. sin 2x/(1 + cos 2x) = tan x

Solution:

Let us solve LHS,

= sin 2x / (1+cos 2x) 

As we know that,

cos 2x = 1 – 2 sin2x

= 2 cos2x – 1

sin 2x = 2 sin x cos x

So,

sin 2x / (1 + cos2x) = [2 sin x cos x / (1 + (2cos2x – 1))]

= [2 sin x cos x / (1+2cos2x – 1)]

= [2 sin x cos x/2 cos2x]

= sin x/cos x

= tan x

LHS = RHS

Hence proved.

Question 4. \sqrt{2+\sqrt{2+cos4x}}=2cosx , 0 < r < π/4

Solution:

Let us solve LHS,

\sqrt{2+\sqrt{2+cos4x}}=\sqrt{2+\sqrt{2+2(2cos^22x-1)}}

As we know that,

 cos 2x = 2 cos2x – 1 ⇒ cos 4x = 2 cos22x – 1

So, 

= \sqrt{2+\sqrt{2+4cos^22x-2}}

= \sqrt{2+\sqrt{4cos^22x}}

= \sqrt{2+2cos2x}

= \sqrt{2+\sqrt{2(2cos^2x-1)}}

= \sqrt{2+4cos^2x-2}

= \sqrt{4cos^2x}

= 2 cos x

LHS = RHS

Hence proved.

Question 5. [1 – cos 2x + sin 2x]/[1 + cos 2x + sin 2x] = tan x

Solution: 

Let us solve LHS,

= [1 – cos 2x + sin 2x]/[1 + cos 2x + sin 2x] 

As we know that,

cos 2x = 1 – 2 sin2x

= 2 cos2x – 1

sin 2x = 2 sin x cos x

So,

= {1 – (1 – 2sin2x) + 2sinxcosx} / {1 + (2 cos2x – 1) + 2 sin x cosx}

= {1 − 1 + 2sin2x + 2sinxcosx} / {1 + 2cos2x − 1 + 2sinx cosx}

= {2 sin2x + 2sinxcosx} / {2 cos2x + 2sinxcosx}

= {2sinx (sinx + cosx)} / {2 cos x (cosx + sin x)}

= sinx/cosx

= tan x

LHS = RHS

Hence proved.

Question 6. [sin x + sin 2x]/[1 + cos x + cos2x] = tanx

Solution:

Let us solve LHS,

= [sin x + sin 2x]/[1 + cos x + cos 2x]

As we know that,

cos 2x = cos2x sin2x

sin 2x = 2 sin x cos x

So,

{sin x + sin 2x} / {1 + cos x + cos 2x} = {sin x + 2 sin x cos x} / {1 + cosx + (2cos2x − 1)}

= {sinx + 2 sinx cos x} / {1 + cosx + 2cos2x − 1}

= {sin x + 2 sin x cosx} / {cosx + 2cos2x}

= {sinx (1 + 2 cos x)} / {cosx(1 + 2cosx)}

= sinx / cosx

= tan x 

LHS = RHS

Hence proved.

Question 7. cos 2x / (1+ sin 2x) = tan (π/4 – x)

Solution:

Let us solve LHS,

= cos 2x / (1 + sin 2x)

As we know that,

cos 2x = cos2x – sin2x

sin 2x = 2 sin x cos x

So,

{cos 2x} / {1 + sin 2x} = {cos2x – sin2x} / {1 + 2 sin x cos x}

= {(cosx – sinx)(cosx + sinx)} / {sin2x + cos2x + 2 sin x cos x}

Since, a2 – b2 = (a – b)(a + b) and sin2x + cos2x = 1

So, 

= {(cosx – sinx)(cosx + sinx)} / {(sinx + cos x)2

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

So, 

 = {(cosx – sinx)(cosx + sinx)} / {(sinx + cosx)(sinx + cosx)}

= (cosx – sinx) / (sin x + cos x)

Now multiplying numerator and denominator by 1/√2, we get,

= \frac{\frac{1}{√2}(cosx-sinx)}{\frac{1}{√2}(sinx+cosx)}

= \frac{(\frac{1}{√2}.cosx-\frac{1}{√2}.sinx)}{(\frac{1}{√2}.sinx+\frac{1}{√2}.cosx)}

= \frac{sin(\frac{π}{4})cosx-cos(\frac{π}{4})sinx}{sin(\frac{π}{4})sinx+cos(\frac{π}{4})cosx}

Since, 1/√2 = sin π/4, so

= \frac{sin(\frac{π}{4}-x)}{cos(\frac{π}{4}-x)}

By using the formulas, we get

sin(A – B) = sinA cosB – sinB cosA 

cos(A – B)= cosA cosB + sinA sinB

= tan (π/4 – x)

LHS = RHS

Hence proved.

Question 8. cos x/(1 – sin x) = tan (π/4 + x/2)

Solution:

Let us solve LHS,

= cos x/(1 – sin x)

As we know that,

cos 2x = cos2x – sin2x

cos x = cos2 x/2 – sin2 x/2 

sin 2x = 2 sin x cos x

sin x = 2 sin x/2 cos x/2

So,

\frac{cosx}{1-sinx}=\frac{cos^2(\frac{x}{2})-sin^2(\frac{x}{2})}{1-2sin(\frac{x}{2})cos(\frac{x}{2})}

= \frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{sin^2(\frac{x}{2})+cos^2(\frac{x}{2})+2sin(\frac{x}{2})cos(\frac{x}{2})}

By using the formulas, 

a2 – b2 = (a – b)(a + b) and sin2x + cos2x = 1), we get

= \frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{[sin(\frac{x}{2})+cos(\frac{x}{2})]^2}

= \frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{[sin(\frac{x}{2})+cos(\frac{x}{2})][sin(\frac{x}{2})+cos(\frac{x}{2})]}

= \frac{cos(\frac{x}{2})-sin(\frac{x}{2})}{sin(\frac{x}{2})+cos(\frac{x}{2})}

= \frac{cos(\frac{x}{2})+sin(\frac{x}{2})}{sin(\frac{x}{2})-cos(\frac{x}{2})}

Now multiply numerator and denominator by 1/√2, we get,

= \frac{\frac{1}{√2}[cos(\frac{x}{2})+sin(\frac{x}{2})]}{\frac{1}{√2}[sin(\frac{x}{2})-cos(\frac{x}{2})]}

= \frac{(\frac{1}{√2})cos(\frac{x}{2})+(\frac{1}{√2})sin(\frac{x}{2})}{(\frac{1}{√2})sin(\frac{x}{2})-(\frac{1}{√2})cos(\frac{x}{2})}

= \frac{sin(\frac{π}{4})cos(\frac{x}{2})+cos(\frac{π}{4})sin(\frac{x}{2})}{sin(\frac{π}{4})sin(\frac{x}{2})-cos(\frac{π}{4})cos(\frac{x}{2})}

= \frac{sin(π/4-x)}{cos(π/4-x)}

 = tan (π/4 – x)

LHS = RHS

Hence proved.

Question 9. cos2 π/8 + cos2 3π/8 + cos2 5π/8 + cos2 7π/8 = 2

Solution:

Let us solve LHS,

= cos2 π/8 + cos2 3π/8 + cos2 5π/8 + cos2 7π/8

As we know that,

cos 2x = 2cos2x – 1

cos 2x+1=2cos2 x

cos2x = (Cos 2x + 1)/2

So,

= cos2 π/8 + cos2 3π/8 + cos2 5π/8 + cos2 7π/8

= \frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{6π}{8})}{2}+\frac{1+cos(\frac{10π}{8})}{2}+\frac{1+cos(\frac{14π}{8})}{2}

= \frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(π-\frac{2π}{8})}{2}+\frac{1+cos(π+\frac{2π}{8})}{2}+\frac{1+cos(2π-\frac{2π}{8})}{2}

= \frac{1+cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}

As we know that, cos (π – A) =- cos A, cos (π+ A) = -cos A and cos (2π – A) = cos A

= 2 x {1 + cos(2π/8)/2} + 2 x {1 – cos(2π/8)/2}

= 1 + cos(2π/8) + 1 – cos(2π/8)

= 2 

LHS = RHS

Hence Proved.

Question 10. sin2 π/8 + sin2 3π/8 + sin2 5π/8 + sin2 7π/8

Solution:

Let us solve LHS,

= sin2 π/8 + sin2 3π/8 + sin2 5π/8 + sin2 7π/8

As we know that, 

cos 2x = 1 – 2sin2x

2sin2x = 1 – cos 2x

sin2x = (1 – cos 2x)/2

So,

= \frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{6π}{8})}{2}+\frac{1-cos(\frac{10π}{8})}{2}+\frac{1-cos(\frac{14π}{8})}{2}

= \frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(π-\frac{2π}{8})}{2}+\frac{1-cos(π+\frac{2π}{8})}{2}+\frac{1-cos(2π-\frac{2π}{8})}{2}

= \frac{1-cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}

As we know, cos (π – A) = -cos A, cos (π + A) = -cos A and cos (2π – A) = cos A

= 2 x {1 – cos(2π/8)/2} + 2 x {1 + cos(2π/8)/2}

= 1 – cos(2π/8) + 1 + cos(2π/8)

= 2

LHS = RHS

Hence proved.

Question 11. (cos α + cos β)2 + (sin α + sin β)2 = 4 cos2(α – β)/2

Solution:

Let us solve LHS,

= (cos α+ cos β)2+ (sin α+ sin β)2

On expanding, we get,

= cos2α + cos2 β + 2 cos α cos β + sin2α+ sin2β + 2 sin α sin β

= 2+2 cos α cos β + 2 sin α sin β

= 2 (1+ cos α cos β+ sin α sin β)

= 2 (1 + cos (α – β))                        [Using, cos (A – B) = cos A cos B+ sin A sin B]

= 2 (1 + 2 cos2(α – β)/2 – 1)            [Using, cos2x = 2cos2x – 1]

= 2 (2 cos2(α – β)/2)

= 4 cos2(α – β)/2

LHS = RHS

Hence Proved.

Question 12. sin2(π/8 + x/2) – sin2(π/8 – x/2) = 1/√2 sin x

Solution:

Let us solve LHS,

= sin2(π/8 + x/2) – sin2(π/8 – x/2)

As we know that, 

sin2A – sin2B = sin (A + B) sin (A-B)

So,

sin2(π/8 + x/2) – sin2(π/8 – x/2) = sin (π/8 + x/2 + π/8 – x/2) sin (π/8 + x/2 – (π/8 – x/2))

= sin (π/8 + π/8) sin (π/8 + x/2 – π/8 + x/2)

= sin π/4 sin x

= 1/√2 sin x                       [As we know that, π/4 = 1/√2]

LHS = RHS

Hence proved.

Question 13. 1 + cos22x = 2 (cos4x + sin4x)

Solution:

Let us solve LHS,

= 1 + cos22x

As we know that, 

cos2x = cos2x sin2x

cos2x+ sin2x = 1

So,

1 + cos22x = (cos2x+ sin2x)2 + (cos2x – sin2x)2 

= (cos4x + sin4x + 2 cos2x sin2x) + (cos4x + sin4x – 2cos2x sin2x)

= cos4x + sin4x + cos4x + sin4x

= 2 cos4x + 2 sin4x

= 2 (cos4x + sin4x)

LHS = RHS

Hence proved.

Question 14. cos32x + 3 cos 2x = 4 (cos6x – sin6x)

Solution:

Let us solve RHS,

= 4 (cos6 x – sin6 x)

On expanding, we get,

4 (cos6 x – sin6 x) = 4 [(cos2x)3 – (sin2x)3]

= 4 (cos2x – sin2x) (cos4x + sin4x + cos2x sin2x)

Now, using the formula, we get

a3 – b3 = (a – b) (a2 + b2 + ab)

= 4 cos 2x (cos4x + sin4x + cos2x sin2x + cos2x sin2x – cos2x sin2x

As we know that, 

cos 2x = cos2x – sin2x

So,

= 4 cos 2x (cos4x + sin4x + 2 cos2x sin2x – cos2x sin2x)

= 4 cos 2x [(cos2x)2 + (sin2x)2 + 2 cos2x sin2x – cos2x sin2x]

By using formula, 

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

= 4 cos 2x [(1)2 – 1/4 (4 cos2 x sin2 x)]

= 4 cos 2x [(1)2-1/4 (2 cos x sin x)2] 

Since

sin 2x = 2sin x cos x

= 4 cos 2x [(12) – 1/4 (sin 2x)2]

= 4 cos 2x (1 – 1/4 sin2 2x)

Since

sin2 x = 1 – cos2x

= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]

= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]

= 4 cos 2x [3/4 + 1/4 cos2 2x]

= 4 (3/4 cos 2x + 1/4 cos³ 2x)

= 3 cos 2x + cos3 2x

= cos3 2x + 3 cos 2x

LHS = RHS

Hence proved.

Question 15. (sin 3A + sin A) sin A + (cos 3A – cos A) cos A = 0

Solution:

Let us solve LHS,

= (sin 3A + sin A) sin A + (cos 3A – cos A) cos A

= (sin 3A) (sin A) + sin2A + (cos 3A) (cos A) – cos2A

= [(sin 3A) (sin A) + (cos 3A) (cos A)] + (sin2A – cos2A)

= [(sin 3A) (sin A) + (cos 3A) (cos A)] – (cos2A – sin2A)

= cos (3A – A) – cos 2A

As we know that, 

cos 2x = cos2A – sin2A

cos A cos B + sin A sin B = cos(A – B)

So,

= cos 2A – cos 2A

= 0

LHS = RHS

Hence Proved.



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Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.1 | Set 2

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      Measurement of angles is a fundamental concept in geometry that deals with quantifying the amount of rotation between two intersecting lines or rays. In Chapter 4 of RD Sharma's Class 11 mathematics textbook, students are introduced to various aspects of angle measurement, including different units
      13 min read
    • Class 11 RD Sharma Solutions - Chapter 4 Measurement of Angles - Exercise 4.1 | Set 2
      Question 11. A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25o in a distance of 40 meters. Solution: Let AB denote the given rail-road. We are given ∠AOB = 25o. We know 180o = π radians = πc or 1o = (π/180)c Hence 25o = 25 × π/180
      6 min read

    Chapter 5: Trigonometric Functions

    • 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 i
      5 min read
    • Class 11 RD Sharma Solutions - Chapter 5 Trigonometric Functions - Exercise 5.1 | Set 2
      Question 14. Prove that [Tex]\frac{(1+cotθ+tanθ)(sinθ-cosθ)}{sec^3θ-cosec^3θ}=sin^2θcos^2θ [/Tex] Solution: We have [Tex]\frac{(1+cotθ+tanθ)(sinθ-cosθ)}{sec^3θ-cosec^3θ}=sin^2θcos^2θ [/Tex] Taking LHS = [Tex]\frac{(1+cotθ+tanθ)(sinθ-cosθ)}{sec^3θ-cosec^3θ}[/Tex] = [Tex]\frac{(1+\frac{cosθ}{sinθ}+\fr
      7 min read
    • Class 11 RD Sharma Solutions - Chapter 5 Trigonometric Functions - Exercise 5.2
      Question 1. Find the values of the other five trigonometric functions in each of the following:(i) cot x = 12/5, x in quadrant III(ii) cos x = -1/2, x in quadrant II(iii) tan x = 3/4, x in quadrant III(iv) sin x = 3/5, x in quadrant I Solution: (i) cot x = 12/5, x in quadrant III As we knew that tan
      7 min read
    • Class 11 RD Sharma Solutions - Chapter 5 Trigonometric Functions - Exercise 5.3
      Question 1. Find the values of the following trigonometric ratios:(i) sin 5π/3 Solution: We have, sin 5π/3 =sin (2π-π/3) [∵sin(2π-θ)=-sinθ] =-sin(π/3) = - √3/2 (ii) sin 17π Solution: We have, sin 17π ⇒sin 17π=sin (34×π/2) Since, 17π lies in the negative x-axis i.e. between 2nd and 3rd quadrant =sin
      8 min read

    Chapter 6: Graphs of Trigonometric Functions

    • Class 11 RD Sharma Solutions - Chapter 6 Graphs of Trigonometric Functions - Exercise 6.1
      The Graphs of trigonometric functions provide a visual representation of how trigonometric functions behave over different intervals. Understanding these graphs is crucial for solving trigonometric equations and analyzing periodic phenomena. This chapter covers the essential techniques to sketch and
      4 min read
    • Class 11 RD Sharma Solutions - Chapter 6 Graphs of Trigonometric Functions - Exercise 6.2
      Question 1: Sketch the following graphs:(i) y = cos (x+[Tex]\frac{\pi}{4}[/Tex]) Solution: To obtain this graph y-0 = cos (x+[Tex]\frac{\pi}{4} [/Tex]), Shifting the origin at [Tex](-\frac{\pi}{4},0) [/Tex], we have X = x+[Tex]\frac{\pi}{4} [/Tex] and Y = y-0 Substituting these values, we get Y = co
      3 min read
    • Class 11 RD Sharma Solutions - Chapter 6 Graphs of Trigonometric Functions - Exercise 6.3
      Sketch the graphs of the following functions:Question 1: y = sin2 x Solution: As we know that, y = sin2 x = [Tex]\frac{1- cos 2x}{2} = \frac{1}{2} - \frac{cos 2x}{2}[/Tex] [Tex]y - \frac{1}{2} = -\frac{cos 2x}{2}[/Tex] On shifting the origin at (0, 1/2), we get X = x and Y = [Tex]y – \frac{1}{2}[/Te
      5 min read

    Chapter 7: Trigonometric Ratios of Compound Angles

    • Class 11 RD Sharma Solutions - Chapter 7 Trigonometric Ratios of Compound Angles - Exercise 7.1 | Set 1
      Question 1. If sin A = 4/5 and cos B = 5/13, where 0 < A, B < π/2, find the values of the following: (i) sin (A + B) (ii) cos (A + B) (iii) sin (A - B) (iv) cos (A - B) Solution: Given that sin A = 4/5 and cos B = 5/13 As we know, cos A = (1 - sin2A) and sin B = (1 - cos2B), where 0 <A, B
      15+ min read
    • Class 11 RD Sharma Solutions - Chapter 7 Trigonometric Ratios of Compound Angles - Exercise 7.1 | Set 2
      Chapter 7 of RD Sharma's Class 11 Mathematics textbook focuses on "Trigonometric Ratios of Compound Angles." This chapter delves into the trigonometric identities related to the compound angles, offering a deeper understanding of how angles interact within the trigonometric functions. It provides es
      13 min read
    • Class 11 RD Sharma Solutions - Chapter 7 Trigonometric Ratios of Compound Angles - Exercise 7.2
      Question 1: Find the maximum and minimum values of each of the following trigonometrical expressions:(i) 12 sin x – 5 cos x(ii) 12 cos x + 5 sin x + 4(iii) 5 cos x + 3 sin (π/6 – x) + 4 (iv) sin x – cos x + 1 Solution: As it is known the maximum value of A cos α + B sin α + C is C + √(A2 +B2), And t
      7 min read

    Chapter 8: Transformation Formulae

    • Class 11 RD Sharma Solutions - Chapter 8 Transformation Formulae - Exercise 8.1
      It is vital for the students working on Class 11 Mathematics to get a hold of the Transformation Formulae. Therefore, the formulae featured in Chapter 8 of RD Sharma’s book serve as a prerequisite for solving different trigonometric problems. This guide will lead you to Exercise 8. 1, with clear spe
      15+ min read
    • Class 11 RD Sharma Solutions - Chapter 8 Transformation Formulae - Exercise 8.2 | Set 1
      In Class 11 Mathematics, Chapter 8 of RD Sharma's textbook focuses on Transformation formulas which are crucial for simplifying complex trigonometric expressions. This chapter helps students understand how to transform trigonometric functions into more manageable forms using specific formulae. Maste
      15+ min read
    • Class 11 RD Sharma Solutions - Chapter 8 Transformation Formulae - Exercise 8.2 | Set 2
      Question 11. If cosec A + sec A = cosec B + sec B, prove that tan A tan B =[Tex]cot\frac{A+B}{2} [/Tex]. Solution: We have, cosec A + sec A = cosec B + sec B => sec A − sec B = cosec B − cosec A =>[Tex]\frac{1}{cosA}-\frac{1}{cosB}=\frac{1}{sinB}-\frac{1}{sinA}[/Tex] =>[Tex]\frac{cosB-cosA}
      4 min read

    Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

    • Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.1 | Set 1
      Prove the following identities: Question 1. √[(1 - cos2x)/(1 + cos2x)] = tanx Solution: Let us solve LHS, = √[(1 - cos2x)/(1 + cos2x)] As we know that, cos2x =1 - 2 sin2x = 2 cos2x - 1 So, = √[(1 - cos2x)/(1 + cos2x)] = √[(1 - (1 - 2sin2x))/(1 + (2cos2x - 1))] = √(1 - 1 + 2sin2x)/(1 + 2cos2x - 1)1 =
      10 min read
    • Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.1 | Set 2
      Prove the following identities: Question 16. cos2 (Ï€/4 - x) - sin2 (Ï€/4 - x) = sin 2x Solution: Let us solve LHS, = cos2 (Ï€/4 - x) - sin2(Ï€/4 - x) As we know that, cos2 A - sin2 A = cos 2A So, = cos2 (Ï€/4 - x) sin2 (Ï€/4 - x) = cos 2 (Ï€/4 - x) = cos (Ï€/2 - 2x) = sin 2x [As we know that, cos (Ï€/2 - A)
      11 min read
    • Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.1 | Set 3
      Chapter 9 of RD Sharma’s Class 11 Mathematics textbook focuses on the trigonometric ratios of the multiple and submultiple angles. This chapter explores the concepts of the trigonometric functions evaluated at angles that are multiples or submultiples of the standard angles. It provides formulas and
      9 min read
    • Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.2
      Chapter 9 of RD Sharma's Class 11 Mathematics textbook delves into the advanced concepts of the trigonometry specifically focusing on the trigonometric ratios of the multiple and submultiple angles. This chapter is crucial for students as it builds on the foundational knowledge of the trigonometric
      7 min read
    • Class 11 RD Sharma Solutions - Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles - Exercise 9.3
      Prove that:Question 1. sin2 72o – sin2 60o = (√5 – 1)/8 Solution: We have, L.H.S. = sin2 72o – sin2 60o = sin2 (90o–18o) – sin2 60o = cos2 18o – sin2 60o = [Tex]\left(\frac{\sqrt{10+2\sqrt{5}}}{4}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2[/Tex] = [Tex] \frac{10 + 2\sqrt{5}}{16} – \frac{3}{4}[/Tex]
      4 min read
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