NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions
NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions are very important for improving maths skills, to solve these questions everybody is required to remember all identities based on trigonometric functions in tips. Here NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions is presented by Future Study Point which is an associate of Future Point Coaching Center a leading publisher of e-books in Amazon. Solutions of all questions of exercise 3.3 are solved and explained by the maths expert, we are sure every student of class 11 th definitely will like and share it.
NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions
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Q1.Prove that
Answer. L.H.S
Hence, proved
Q2.Prove that
Answer.
L.H.S
Since, Cosec (π+θ) = – cosecθ
Hence, proved
Q3.Prove that
Answer.
L.H.S
5π/6 = (5×180)/6=150 =180° –30°= π – π/6(π rad.=180°)
Substituting cotπ/6 =√3, tanπ/6 =1/√3
3 + 2 + 1 =6 = R.H.S
Q4. Prove that
Answer. L.H.S
Substituting 3π/4 = (3×180)/4 = 135=180°–45° =π – π/4
= 1 + 1 + 8 =10= R.H.S
Q5. Find the value of
(i) sin75°
(ii) tan15°
Answer.
(i) sin75° = sin(45 +30)
sin(45 +30) = sin45.cos30 + cos45.sin30 [sin(x +y) = sinx.cosy + cosx.siny]
(ii) tan15° = tan(45 –30)
Multiplying denominator and numerator by (√3 –1)
=2– √3
Therefore, tan 15° = 2 – √3
Q6. Prove that
Answer.
L.H.S
Multiplying and dividing R.H.S by 2.
Applying the identities, 2cosA.cosB = cos(A +B) +cos(A –B)
and –2sinA.sinB = cos(A+B) – cos(A–B)
= sin(x +y)= R.H.S
Q7. Prove that
Answer.
L.H.S
Applying the following identities
Q8. Prove that
Answer.
L.H.S
As we know cos(π+x) = –cosx, cos(–x) =cosx, sin(π–x) =sinx, cos(π/2 +x)=–sinx
= cot²x = R.H.S
Q9.Prove that
Answer.
L.H.S
3π/2 +x = 3×180/2+x =270+x=180+90 +x= π + π/2+x
cos(3π/2+x) = cos[π +(π/2 +x)] = –cos(π/2 +x)=–(–sinx)=sinx,cos(2π+x)=cosx
Similarily,cot(3π/2+x)= tanx and cot(2π +x) = cotx
=sinx.cosx(tanx + cotx)
= 1= R.H.S
Q10. Prove that: sin(n +1)x.sin(n+2)x +cos(n+1)x.cos(n+2)x = cosx
Answer. L.H.S sin(n +1)x.sin(n+2)x +cos(n+1)x.cos(n+2)x
Multiplying and dividing it by 2
= cosx
Q11. Prove that:
Answer.
L.H.S
3π/4 = 3 ×180/4 = 135 = 180 –45 = π – π/4
sin(π – θ) = sinθ, so
Q12.Prove that sin²6x – sin²4x = sin2x.sin10x
Answer.
L.H.S
sin²6x – sin²4x
=(sin6x + sin4x)(sin6x – sin4x)
=2sin5x.cosx.2cos5x.sinx
=2sin5x.cos5x. 2sinx.cosx
=sin10x.sin2x=R.H.S
Q13.Prove that
cos²2x – cos²6x = sin4x.sin8x
Answer
L.H.S
cos²2x – cos²6x
=(cos2x +cos6x)(cos2x – cos6x)
=2cos4x.cos(–2x).[–2sin4x.sin(–2x)]
sin(–θ) = – sinθ, so
= (2sin4x.cos4x).(2sin2x.cos2x)
=sin8x.sin4x = R.H.S
Q14. Prove that sin2x + 2sin4x + sin6x = 4cos²x.sin4x
Answer.
L.H.S
sin2x + 2sin4x + sin6x
(sin2x + sin6x )+ 2sin4x
= 2sin4x.cos(–2x) + 2sin4x
= 2sin4x.cos2x + 2sin4x
=2sin4x(1 + cos2x)
As we know cos2x = 2cos²x – 1,so
=2sin4x.2cos²x
=4cos²x.sin4x = R.H.S
Q15. Prove that cot4x(sin5x + sin3x) = cotx(sin5x – sin3x)
Answer.
L.H.S
cot4x(sin5x + sin3x)
=cot4x[2sin4x.cosx]
=2.cos4x.cosx
For getting cotx in R.H.S ,dividing and multiplying above expression by sinx
As we know, 2cosA.sinB = sin(A +B) – sin(A –B)
=cotx(sin5x – sin3x) = R.H.S
Q16. Prove that
Answer.
L.H.S
Applying the following identities
Q17. Prove that
Answer.
L.H.S
Applying the following identities
= tan4x
Q18. Prove that
Answer.
L.H.S
Applying the identities
Q19. Prove that
Answer.
L.H.S
Applying the following identities
=tan2x
Q20.Prove that
Answer.
L.H.S
Applying the following identities
And cos²x – sin²x = cos2x
= –2 × –sinx
=2sinx = R.H.S
Q21.Prove that
Answer.
L.H.S
Arranging the term as follows
Applying the following trigonometric identities
cosA +cosB = 2cos(A+B)/2.cos(A–B)/2 and sinA + sinB = 2sin(A+B)/2.cos(A–B)/2
= cot3x = R.H.S
Q22.Prove that cotx.cot2x – cot2x.cot3x – cot3x.cotx= 1
Applying the following trigonometric identity
cot3x.cotx + cot3x.cot2x = cot2x.cotx – 1
cotx.cot2x –cot2x.cot3x – cot3x.cotx = 1
L.H.S = R.H.S
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