Cbse class 11 maths chapter 3 solutions pdf

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Cbse class 11 maths chapter 3 solutions pdf

Answer: (1) We know that radiating so, (we need to take) (we use and 1 '= 60' ') here 1' represents 1 minute and 60 "represents 60 seconds now, (ii) -4 we know that radians (we need to take) So, -4 = radians (we use and 1 '= 60' ') (iii) we know that the radiation (we need to take) Then, (iv) we know that radiant (we need to take) So, question: 2 find the values of the other five trigonometric functions x is in the second quadrant answer :. Solution x is in the second quadrant Therefore cosec x x is positive lies in the second quadrant, therefore cos x is negative x lies .. . In the second quadrant therefore sec x is x negative lies in the second quadrant therefore tan x is x negative lies in the second quadrant therefore cradle x is a negative question: .. 1 Demonstrate what answer: We know that the values of sin (30 degrees ), cos (60 degrees) and tan (45 degrees) = Ie :. RHS question: 2 Prove that response: the solutions to the proble But given as follows RHS. Question: 3 Demonstrate what answer: We know the values of C OT (30 degrees), Tan (30 degrees) and Cosec (30 degrees) R.H.S. Question: 4 Demonstrate what answer: using the values mentioned above R.H.S. Question: 5 (i) find the response value: we know that (sin (x + y) = sinxcosy + cosxsiny) using this identity formed demand: 6 Demonstrate the following: Answer: Multiply and divide for 2 both so and sinful functions that We now get that we know that 2xacosb = COS (A + B) + COS (AB) - (i) -2Sinasinb = COS (A + B) - COS (AB) - (II) We use these two identities in our question A = B = So, as we know that using this RHS question: 8 Demonstrate the following response: as we know that ,,, and using these our equation simplify the demand RHS: 9 Demonstrate the following answer: We know that therefore, using these our simplifies equation for RHS Question: 10 Demonstrate the following response: Multiply and divide for 2 hours, using the identities -2sinasinb = COS (A + b) - COS (A-B) 2, which cos (A + B) + COS (A-B) R.H.S. Question: 11 Demonstrate the following response: We know that [COS (A + B) - COS (A-B) = -2Sinasinb] Using this identity R.H.S. Question: 12 Demonstrate the following response: We know that so, now, we know that using these Sin6x + sin4x identities = 2Sin5x COSX SIN6X - SIN4X = 2COS5X SINX NOW, 2SinacOSB = SIN (A + B) + SIN (AB) 2BOXASINB = sin (a + b) - sin (ab) using these identity 2cos5x sin5x = sin10x - 0 2sinx cosx = sin2x + 0 then question: 13 demonstrate the following response: as we know what time using these cos2x identity - cos6x = -2sin (4x ) SIN (-2X) = 2SIN4XSIN2X (SIN (-X) = -SIN X COS (-X) = COSX) COS2X + COS 6X = 2COS4XCOS (-2X) = 2COS4XCOS2X So our equation becomes RHS demand: 14 Demonstrate the following Answer: We know that we are using this sin2x + 2Sin4x + sin6x identity = (sin2x + sin6x) + 2sin4x sin2x + sin6x = 2sin4xcos (-2x) = 2sin4xcos (2x) (so (-x) = so x) then, our equation BECOMES SIN2X + 2SIN4X + SIN6X = 2SIN4XCOS (2X) + 2SIN4X NOW, Take the Municipality 2SIN4X SIN2X + 2SIN4X + SIN6X = 2SIN4X (COS2X +1) () = 2SIN4X (+1) = 2SIN4X () = RH S Question: 15 Demonstrate the following response: We know that using this, we get sin5x + sin3x = 2sin4xcosx now nultiply and divide for sin x now we know that with this our equation becomes r.h.s. Question: 16 Demonstrate the following response: as we know that R.H.S. Question: 19 Demonstrate the following response: We know that R.H.S. QUESTION: 22 Demonstrate the following response: cot x cot2x - cot3x (cot2xcotx) Now we can write cot3x = cradle (2x + x) and we know that then, = cotx cot2x - (cot2xcotx -1) = cotx cot2x - cot2xcotx +1 = 1 = RHS question: 23 Demonstrate what answer: We know that and we can write 4x tan tan = 2 (2x) then, = = = = = rhs Question: 24 Demonstrate the following answer: We know that we use this in Problem COS 4x = COS 2 (2x) = = = = R.H.S. Question: 25 Demonstrate the following answer: We know that so 3x = 4 - 3COS X We use this in our problem there can write COS 6x as so 3 (2x) COS 3 (2x) = 4 - 3 COS 2x = - = = 32 - 4 - 48 + 24 - = 32 - 48 + 18 - 1 = RHS Question: 5 Find the general solution for each of the following response Equation: COS4X = COS2X COS4X - COS2X = 0 0 know we use this identity cos 4x - 2x = cos -2sin3xsinx -2sin3xsinx sin3xsinx = 0 = 0 Therefore, with this it can sin3x sinx = 0 or 0 = 3x = x = x = x = Therefore, the general solution is the general solution Question is: 1 Demonstrate that answer: we know that cos a + b = cos we use this in our problem (we know that cos (-x) = cos x) Continue using the identity above, we know SO = 0, = 0 = RHS Question: 2 Try one answer: we know and we use it in our problem = + = (4SINX - 4) + sin x (4 - 4cos X) cosx now take the common 4SINX by 1 ? ? term -4Cosx from 2 ? st term = 4 (1 -) - 4 (1 -) = 44 = 0 = RHS Question: 3 Try one answer: we know and we use these two in our problem and = + = 1 + 2Cosxcosy + 1 - 2SINXSINY = 2 + 2 (CosxCosy - SinxSiny) = 2 + 2cos (x + Y) = 2 (1 + cos (x + y)) Now we can write = = = RHS Question: 4 Try one answer: we know and we use these two in our problem = + = 1 - 2cos xcosy + 1 - 2sinxsiny = 2-2 (cosxcosy sinxsiny +) = 2 - 2cos (x - y) = 2 (1 - cos (x - y)) Now we can write so = RHS Question: 5 Prove that answer: we know we use this identity to our problem if we see that we need to SIN4X in our final result, so it's better if we had a combination of SIN7X and SIN X, SIN3X and SIN5X TP Get Sin4x take 2SIN4X Common = 2SIN4X (Cos3x + cosx) we know now that we use this = = 2sin4x () = 4Cosxcos2xsin4x = RHS Question: 6 Try one answer: we know and we use these two identities in our problem SIN7X + SIN5X = = SIN 9x + SIN 3X = = cos 7x + cos5x = = cos cos3x 9x + = = = = = RHS Question: 7 Prove that answer: we know that we use these identities sin2x 2SINX cosx + = + = take 2 sinx Common r.h.s. Question: Find 8 in x in quadrant II Answer: We know that tan x = = = x is in the second quadrant that's why there is sec x So, now, we know that = () = = = = = = x lies in Quadrant II so worth iS + VE = We know that 1 = - = x lies in Quadrant II so the SIN value is x + VE Question: 9 Find where, x in Quadrant III Answer: We know that cos x = cos x + 1 + 1 = = = Now, we know that cos x = 1 = - = = ? Why is + ve on demand dial: Find 10 in x in quadrant II response: all functions they are positive in this range ? ? what we know 1 = - = = cos x = (cos x is -ve in quadrant II) We know that cos x = (? because all functions are posititve within a certain range) at Similarly, X = cos (? because all functions are posititve in a given interval) NCERT solutions for the class 11 mathematics NCERT solutions for class 11- Subject essay the basic identity used in NCERT solutions for the class mathematics 11 Ca ap te r 3 The trigonometric functions are listed below the identities of the above is possible that it was designed in your high school classes. Here are a couple of identity that you have to remember and learn from the NCERT solutions for class 11 Chapter 3 mathematical functions trigonometric Some identity conditioned by NCERT solutions for class 11 mathematics Chapter 3 trigonometric functions If angles x, y (x ? ? ? y ) is not an odd multiple of ? ? ? 2, then if the angles x, y (x ? ? y) is not a multiple of ? ? ?, so there are some other identity used in NCERT solutions for chapter 11 Maths class 3 trigonometric functions that can be derived using the identity above. Try to derive it from yourself. Enjoy the reading !!! NCERT Solutions for Class Maths Chapter 11 3 3.1 EX trigonometric functions are part of NCERT solutions for math class 11. Here we gave NCERT solutions for Maths Class 11 Chapter 3 Functions trigonometric EX 3.1. Council CBSE NCERT Text Book of Class 11 Class Subject mathematical functions Chapter 3 Chapter Chapter Name Number of Exercise trigonometric Ex 3.1 Solved 7 Category NCCT Solutions necert Solutions for class 11 Mathematical functions Chapter 3 Trigonometric ex 3.1 ex 3.1 Class 11 maths?, question 1. Find the radiant measures corresponding to the following degree measures: (i) 25 ? ? (ii) -47 ? ? 30 Write note (iii) 240 ? ? (iv) 520 ? ? Solution. We, 180 ? ? = ? ?,? ?,? ?,? 3.1 Class 11 Question Maths 2. Find the degree measures corresponding to the Radial measures (i) (ii) -4 (iii) (iv) solution. We ? ?,? ?,? radians = 180 ? ? ex 3.1 class 11 mathematical question 3. A wheel makes 360 rpm in one minute. Through how many Ramians is addressed in a second? Solution. Number of revolutions made in wheel in a minute = 360 as we know that, 1 revolution = 27 ? ?,? ?,? radians ?,360 rpm = 720 ? ?,? ?,? ?,? radians ? ?In 1 minute wheel can do = 720 ? ?,? radians ?, ? ?in 60 seconds the wheel can do = 720 ? ?,? ?,? radians ?, ?, ?, ?, ?, ? ?in 1 second wheel can make the ex 3.1 class 11 maths-question 4. Find the measurement of the degree of the subordinate angle in the center of a radius circle 100 cm from a bow of a length 22 cm solution. Let the center and AB is the length of the circle bow. Ex 3.1 class 11 maths question 5. In a circle of diameter 40 cm, the length of an agreement is 20 cm. Find the length of the minor Xhe Chord arc. Solution. Let it be the smaller bow of the rope. AB = 20 cm, OA = OB = 20 cm ex 3.1 Class 11 Mathematical question 6. If in two circles, the arcs of the same length subtend angles from 60 ? ? and 75 ? ? in the center, find the relationship between them, radii. Solution. Let R1 R2 and ?z?1, ?z?2 are the radii and the angles subordinated respectively in the center of two circles. Ex 3.1 class 11 maths 1. Question 7. Find the radian angle through which a pendulum oscillates if its length is 75 cm and the tip describes a bow length (i) 10 cm (ii) 15 cm ( Iii) 21 cm solution. We hope that necert solutions for class 11 maths chapter 3 trigonometric functions ex 3.1 helped. If you have any questions regarding necert solutions for class 11 maths chapter 3 trigonometric functions ex 3.1, releases a comment below and we will reply to the sooner. Early.

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