Class 9 : Trigonometry 2

1. Find the value of sin(-110^o)
Solution: Here,

     Sin(-110) = - sin110^o

                     = -sin(90^o+30^o)

                     = - cos30^o

                    = - √32 .


2. Find the value of cos(-690^o).
Solution: Here,     Cos(-690^o) = cos690^o
                                = cos(2⨉360^o-30^o)
                                = cos30°                               = √3/2


3.  Find the value of  tan1020^o.Solution: Here,

        tan1020° = tan(3⨉360° – 60°)

                       = tan60^o
                       = √3.

4. Prove that: sin65° +cos35° = cos25° +sin55°
Proof: Here,

LHS = sin65° + cos35°         

        = sin(90°-25°) + cos(90°-55°)

        = cos25° + sin55°

         = RHS.

Proved.

5. Prove that: tan9^o.cot27^o – cot81^o.tan63^o = 0

Proof: Here,

LHS = tan9^o.cot27^o – cot81^o.tan63^o

        = tan(90^o – 81^o).cot(90^o-63^o) – cot81^o.tan63^o

        = cot81^o.tan63^o – cot81^o.tan63^o

        = 0

        = RHS.

Proved.

6. If A,B&C are the angles of triangle, prove that : Sin(A+B) = sinC.

 Proof: Here, A,B&C are the angles of a triangle

So,  A+B+C = 180°

  ⇒ A+B = 180° - C

Operating both sides by sin,we get

   sin(A+B) = sin(180°-C)

⇒ Sin(A+B) = SinC.

Proved.

7. Find the value of A if sin4A = cosA.
Proof: Here,
     sin4A = cosA
⇒ sin4A = cosA
⇒ sin4A = sin(90^o-A)
⇒ 4A = 90^o - A
⇒ 4A+A = 90^o
⇒ 5A=90^o
⇒ A=18^o.
∴   A = 18^o.

8. Find the value of x, if xcos(90°+A).cos(90°-A)+tan(180°-A)cot(90°-A) = sinA.sin(180°-A).
Proof: Here,
       xcos(90+A).cos(90-A)+tan(180-A)cot(90+A) = sinA.sin(180-A).
⇒  -xsinA.sinA +tanA.tanA = sinA.sinA
⇒  -xsin²A + tan²A = sin²A
⇒ sin²A + xsin²A = tan²A
⇒ sin²A( 1+x) = tan²A
⇒ 1+ x = sec²A
⇒ x = sec²A-1`
⇒ x = tan²A.
∴  x = tan²A.

9. If A,B& C are angles of a triangle, prove that:  tan(A+B)+tanC = 0.
Solution: Here,
 A,B& C are angles of a triangle.
So, 
    A+B+C =180^o
i.e.  A+B = 180° - C.
Operating both sides by tan,we get
      tan(A+B) = tan(180-C)
⇒ tan(A+B) = - tanC
⇒ tan(A+B) + tanC = 0
      LHS = RHS 

Proved.

10. If  A =30^o & B=60^o, verify : sin(A+B) = sinAcosB + cosAsinB.Solution: Here, A =30^o & B=60^o. 
     Now, sin(A+B) = sinAcosB + cosAsinB.
⇒ sin(30^o+60^o) = sin30^o.cos60^o+cos30^o.sin60^o
⇒  sin90^o   = 12 .12 + √32.√32    ⇒ 1 = 14 + 34
 ⇒ 1 = 1

Verified.

11. Solve for x ,

xcotA.tan(90°+A) = tan(90°+A).cot(180°-A) +xsec(90°+A).cosecA.

Sokution: Here,

  xcotA.tan(90+A) = tan(90+A).cot(180-A)  

+xsec(90+A).cosecA

⇒  xcotA.(-cotA) = -cotA.(-cotA)+x(-cosecA).cosecA
⇒ -xcot²A = cot²A - xcosec²A
⇒ xcosec²A - xcot²A = cot²A
⇒  x(cosec²A-cot²A) = cot²A.
⇒  x ⨉ 1 = cot²A
⇒  x = cot²A.
∴   x = cot²A.

12. Prove that: sin(90°+A).cos(180°-A).tan(180°-A) = sin²(90°+A).cot(90°-A).

Proof: Here,

  LHS = sin(90°+A).cos(180°-A).tan(180°-A)
           = cosA.(-cosA).(-tanA)
           = cos²A.tanA
RHS =  sin²(90°+A).cot(90°-A).
       = cos²A.tanA
       = LHS.

Proved.

13. Without using trigonometric tables, evaluate sin 35° sin 55° - cos 35° cos 55°.
Solution: Here,
      sin 35° sin 55° - cos 35° cos 55°
   = sin 35° sin (90° - 35°) - cos 35° cos (90° - 35°),

   = sin 35° cos 35° - cos 35° sin 35°,     

[Since sin (90° - θ) = cos θ and cos (90° - θ) = sin θ]

   = sin 35° cos 35° - sin 35° cos 35°
   = 0.

14. If sec 5θ = cosec (θ - 36°), where 5θ is an acute angle, find the value of θ.
Solution: Here,
     sec 5θ = cosecc (θ - 36°)
⇒ cosec (90° - 5θ) = cosec(θ - 36°), [Since sec θ = csc (90° - θ)]
⇒ (90° - 5θ) = (θ - 36°)
⇒ -5θ - θ = -36° - 90°
⇒ -6θ = -126°
⇒ θ = 21°, [Dividing both sides by -6]
Therefore, θ = 21°.

15. Using trigonometrical ratios of complementary angles prove that tan 1° tan 2° tan 3° ......... tan 89° = 1.
Proof: Here,
LHS= tan 1° tan 2° tan 3° ...... tan 89°
       = tan 1° tan 2° ...... tan 44° tan 45° tan 46° ...... tan 88° tan 89°
       = (tan 1° ∙ tan 89°) (tan 2° ∙ tan 88°) ...... (tan 44° ∙ tan 46°) ∙ tan 45°
       = {tan 1° ∙ tan (90° - 1°)} ∙ {tan 2° ∙ (tan 90° - 2°)} ...... {tan 44° ∙ tan (90° - 44°)} ∙ tan 45°
       = (tan 1° ∙ cot 1°)(tan 2° ∙ cot 2°) ...... (tan 44° ∙ cot 44°) ∙ tan 45°,
                               [Since tan (90° - θ) = cot θ]= (1)(1) ...... (1) ∙ 1, [Since tan θ ∙ cot θ = 1 and tan 45° = 1]
       = 1
 Therefore, tan 1° tan 2° tan 3° ...... tan 89° = 1.`

16. Find the value of tan 240°.
Solution: Here,
tan (240)° = tan (180 + 60)°
              = tan 60°;     [since we know tan (180° + θ) = tan θ.]
              = √3.

17. Find the value of sec 210°.
Solution: Here,
       sec (210)° = sec (180 + 30)°
                        = - sec 30°;
[since we know sec (180° + θ) = - sec θ]
                        = -2√3.

18. Find the value of tan 240°.
Solution: Here, tan (240)° = tan (180 + 60)°
              = tan 60°; since we know tan (180° + θ)
                           = tan θ
                           = √3.

19. Find the value of sec 150°.
Solution: Here,
      sec 150° = sec (180 - 30)°

                     = - sec 30°; 

[since we know, sec (180° - θ)   = - sec θ]

                     = - 2√3.

20. Find the value of tan 120°.
Solution: Here,
     tan 120° = tan (180 - 60)°
      = - tan 60°; since we know, tan (180° - θ) = - tan θ
                    = - √3.
21. Find the value of cos 200° sin 160° + sin (- 340°) cos (- 380°).
Solution: Here,
Given,
       cos 200° sin 160° + sin (- 340°) cos (- 380°)
    = cos (2 × 90° + 20°) sin (1 × 90° + 70°) + (- sin 340°) cos 380°
    = - cos 20° cos 70° - sin (3 × 90° + 70°) cos (4 × 90° + 20°)
    = - cos 20° cos 700 - (- cos 70°) cos 20°
    = - cos 200 cos 70° + cos 70° cos 20°
    = 0.

22. Express cos (- 1555°) in terms of the ratio of a positive angle less than 30°.
Solution: Here,
cos(- 1555°) = cos 1555°, 
 [since we know cos (- θ) = cos θ]
                     = cos (17 × 90° + 25°)
                     = - sin 25°;
[since the angle 1555° lies in the second d quadrant and cos ratio is negative in this quadrant. Again, in the angle 1555° = 17 × 90° + 25°, multiplier of 90° is 17, which is an odd integer ; for this reason cos ratio has changed to sin.].

23. A, B, C, D are the four angles, taken in order of a cyclic quadrilateral. Prove that, cot A + cot B + cot C + cot D = 0.
Proof: Here.
We know that the opposite angles of a cyclic quadrilateral are supplementary.
Therefore, by question we have,
      A + C= 180°               or, C = 180° - A;
And B + D= 180°               or, D = 180° - B.

Therefore,
  L. H. S. = cot A + cot B + cot C + cot D
               = cot A + cot B + cot (180° - A) + cot (180° - B)
               = cot A + cot B - cot A - cot B
               = 0.
               = RHS.

Proved.

24. Prove that:

sinA1+cosA + sinA1-cosA  = 2cosecA
Proof: 
LHS = sinA1+cosA + sinA1-cosA 
         = sinA1-cosA+ sinA1+cosA1+cosA1-cosA
         = sinA-sinA.cosA+sinA+sinA.cosA1-cos2A
         = 2sinAsin2A                                [∴ 1 - cos² A = sin² A ]
        = 2sinA
        = 2 cosecA                           [∴ cosecA = 1sinA]
        = RHS.

Proved.

25. 

Find the value of:
cosec (-1170°)

Solution: Here,
cosec (-1170°) 
= - cosec (1170°)
= - cosec (3 × 360° + 90°)
= - cosec 90° 
= -1 
 26. If secθ + tanθ= x, show that sinθ = x2 – 1x2+ 1.Solution: Given:
     secθ + tanθ= x
⇒ 1cosθ + sinθcosθ = x
⇒    1+sinθcosθ = x
Squaring on both sides,
      (1+sinθcosθ)² = x²
⇒    (1+sinθ)2cos2θ = x²⇒      (1+sinθ)21 - sin²θ = x²
⇒    1+sinθ2(1-sinθ)(1+sinθ) = x²
⇒   1+sinθ1-sinθ = x²
⇒   1 + sinθ = x² - x²sinθ
⇒   sinθ + x²sinθ = x² - 1 
⇒   sinθ(1 + x²) = x² - 1 
⇒  sinθ = x2 – 1x2+ 1
∴  sinθ = x2 – 1x2+ 1

Proved.

27. If A = 2B = 3C = 90°, find the value of cos2A – cot2B + cosec2C.
Solution: Here,
A = 90° 
2B = 90° Þ B = (90° )/2 = 45°
3C = 90° Þ C = (90° )/3 = 30°
Now, 
cos²A - cot²B + cosec²C
= cos²90° - cot²45° + cosec²30°
= 0² + (1)² + (2)² 
= -1 + 4 
= 3
 28. If cosA=45, find the values of cosecA and tanA.Solution: We have,
           cosA=45
Now, cosecA=1sinA
                      =1√{1-cos2A}
                      =11-452
                      =5√{52-42
                      =53
Again,
TanA=sinAcosA
         =1cosecA.cosA
        =153.45
        =34.
29. If  tanA= 2aba2-b2, find the values of cosA and sinA.Solution: find your self. If you can't do, ask me
30. If pcotA=√p2-q2, then find the value of sinA.Solution: find your self. If you can't do, ask me

31. If 3cosA=2 then show that 6⁡sin2A=5cosA.Solution:
We have, 3 cosA=2
i.e.  cosA=23
To prove,6sin2A=5cosA
Here, LHS=6sin2A
                   =6(1- cos²A)
                   =6(1-2²3² )
                   =6(1-49 )
                   =103
Again,
        RHS=5cosA
                =5×23
                 =103
Hence,
       LHS=RHS

Proved.

32. 

if A is angle and 5sinA=3, find the value of 4cosecA-3cotA.

Solution: We have,
      5sinA=3
i.e.  sinA=35
Now, 
4cosecA-3cotA
      =4sinA-3cosAsinA
     =4sinA-3√(1 -⁡sin2A)sinA
     =4-3√(1-sin2A)sinA
     =4-3√1-35235
     =83.  
33. If X is an acute angle and
cosecX=135, find the value of 13cosX+24tanX.
Solution: 
We have,
cosecX=135
Now, sinX=1cosecX=513
cosX=√1-sin2X= √1-5132=1213
tanX=sinXcosX=5131213=512
Hence, 13cosX+24tanX
         =13×1213+24×512
         =12+10
         =22.
34. If A is an acute angle and tanA= 34, find the value of 5cosecA+10secA.Solution: 
We have,⁡tanA=34
Now, secA= √1+⁡tan2A=√1+342=54
          cotA=1tanA=43
           cosecA= √1+⁡cot2A=√1+432=53
Now,
         5cosecA+10secA
     =5×53+10×54
      =1256.  
35. Find the value of :⁡sin2135°+cos2120°- ⁡sin2120°+tan2150°

Solution: We have, 

sin2135°+⁡cos2120°-⁡sin2120°+⁡tan2150° 
=⁡sin2(180°-45°)+⁡cos2(180°-60°)-⁡sin2(180°-60°)+ tan²(180°-30°)
=sin245+cos260-⁡sin260+⁡tan230
=1√22+122-√322+1√32
=12+14-34+13
=13. 
36. If 15 sin²θ + 2 cos² θ = 7, find the value of tan θ.Solution: Given,
      15 sin²θ  +  2 cos²θ  = 7
⇒   15 sin²θ + 2 (1 - sin²θ) = 7
⇒  15 sin²θ  + 2 - 2sin²θ) = 7
⇒ 13 sin²θ = 5
⇒ sin²θ = 513
⇒  sinθ = ±√5√13 = ph
∴when p = √5, then h = √13
     b = √h2-P2
        = √13-5
        = √8
        = 2√2.
Finally,
    tanθ = Pb = √52√2. 
37. If 2 + 2 tan θ = sec² θ, find the value of tan θ.Solution: Given,
      2 + 2 tan θ = sec ² θ
⇒  2 + 2 tan θ =1 + tan ² θ
⇒ tan² θ  - 2tan θ  - 1  = 0 ......(i)
Comparing this with ax² + bx + c = 0 
We get a =1, b = - 2 , c = -1 & x = tan θ
∴ x = ⁡tanθ= -b ± b2-4ac2a
       = 2 ± √-22-4×1-12×1
      = 2 ± 2√22
     = 1±√2.
∴ tanθ= 1 ± √2.  
38. If sin A = m and tan A = n,
Prove that: m^-2 - n^-2= 1
Solution: Given , sin A = m
⇒  1m = 1sin A
⇒  cosec A = 1sin A
⇒  cosec A = 1m
⇒  cosec² A = 1m 2 ...........(i)
and tan A = n
⇒  1tan A = 1n
⇒ cotA= 1n
⇒  cot² A = 1n2 .............(ii)
Subtracting (ii) form (i),
cosec²A - cot²A = 1m 2 – 1n 2
⇒ 1 = m^-2 - n^-2            (∵ cosec² A - cot² A = 1 )
∴ m^-2 - n^-2 = 1.

Proved.

39. If ⁡secθ+ tanθ= x, show that:sinθ= x 2- 1x 2+ 1. 

Solution: Given,
        sec θ + tan θ = x
⇒  1cosθ + sin θ⁡cosθ = x
⇒  1+ sin θ⁡cosθ = x
⇒  1+ sin θ = x cos θ
Squaring both sides,
      (1 + sinθ)² = x² cos² θ
⇒  1 + 2sin θ + sin² θ = x² (1 -  sin² θ)
⇒  1+ 2sin θ + sin² θ = x² - x² sin² θ
⇒  (1 + x²) sin² θ + 2 sin θ + (1 - x² ) = 0
⇒  (1 + x²) sin² θ + {(1 + x² ) + (-x² )sin θ} + (1 - x² ) = 0
⇒ (1 + x²) sin² θ + (1 + x² ) sin θ + (1 - x² )sin θ + (1 - x² ) = 0
⇒  (1 + x²) sin θ  (sin θ+1) + (1 - x² ) (sin θ + 1) = 0
⇒ [(1 + x²) sin θ + (1 - x² )] (sin θ + 1) = 0.
Either ,(1 + x²) sin θ + (1 - x²) = 0  
⇒ (1 +  x²) sin θ = x² - 1
⇒  ⁡sinθ= x 2- 1x 2+ 1
∴⁡sinθ= x 2- 1x 2+ 1.  
40. If cos⁴A + cos²A = 1, then prove that: sec⁴A - sec²A = 1
Proof: Given,
      cos⁴ A + cos² A = 1
⇒  1sec 4 A + 1sec 2 A = 1 ∴⁡cosA= 1sec A
⇒  1+sec 2 Asec 4 A = 1 
⇒  1+ sec 2 A = sec 4 A
⇒  1 = sec 4 A – sec 2 A 
∴    sec ⁴ A - sec ² A = 1.

Proved.

41. If cos⁴A + cos²A = 1, then prove that: tan⁴A + tan²A = 1

Proof: Given,

     cos ⁴  A + cos ²  A = 1 
⇒   cos ⁴  A = 1 - cos ²  A
⇒  cos ⁴  A = sin ²  A.
Now, 
LHS = tan ⁴  A + tan ²  A
        = sin 4 Acos 4 A + sin 2 Acos 2 A
       = cos 4 A2cos 4 A + cos 4 A2cos 2 A         (∵ cos ⁴ A = sin² A)
        = cos⁴ A + cos ² A (Given cos ⁴ A + cos ² A = 1)
        = 1.
∴ tan ⁴A + tan ²A = 1.
Proved.
42. If 3 sin B + 5 cos B = 5, show that: 3 cos B - 5 sin B = ± 3.

43. 

If tan A + sin A = p and tan A - sin A = q, prove that:
P² - q² = 4 √pq

Solution: Given,

                (tan A + sin A) = p ............... (i)
                (tan A - sin A) = q................ (ii)
Multiplying   (i) & (ii) 
      tan² A - sin² A = pq
⇒   pq = sin 2 Acos 2 A – sin 2 A
              = sin 2 A-sin 2 A cos 2 Acos 2 A
              = sin 2 Acos 2 A  1- cos 2 A
              = tan ² A. sin ² A
∴ √pq  =tan⁡tanA.sin⁡sinA……………..iii

Again squaring & subtracting (i) & (ii)
(tan A + sin A ) ² = p²............... (i)
(tan A -  sin A) ² = q²................ (ii)
     tan² A+ 2 tan A sin A + sin² A - tan² A + 2 tan A sin A - sin² A = p² - q²
⇒   4 tan A sin A = p² - q²
⇒   4√pq = p² - q² [ ∵ from (iii) ]
∴   p² - q² = 4√pq.

Proved.

44. 

If x sin³ θ - sinθcosθ +y cos³ θ = 0 and tanθ= yx . Show that x² + y² = 1.

Solution: Given,

      tanθ= yx
⇒   sin θcosθ = yx
⇒  sinθ= yx⁡cosθ
Now,
Put sin θ = (y )/x cosθ in
      x sin³ θ - sin² θ cos θ + y cos³ θ = 0
⇒   yx cos θ³ - yx² cos² θ+ y. cos³ θ = 0 
⇒   y 3x 2 cos³ θ - y 2x 2 cos³ θ + y cos³ θ = 0
⇒  y ³ cos³ θ - y ² cos³ θ + x ² y cos³ θ = 0
⇒  y cos³ θ(y ²- 1+ x ²) = 0
⇒  y ²- 1+ x ² = 0
⇒  x ²+ y ² = 1.

∴  LHS = RHS
Proved.

45. If 1sin⁡sinA + 1sin⁡sinB+1sin⁡sinC = 0, then prove that: sin⁡sinA+sin⁡sinB+sin⁡sinC² = 3 - cos² A + cos² B + cos² C

Proof: Given,

1sinA + 1sinB + 1ssinC = 0
⇒  1sinA + 1⁡sinB = -1⁡sinC
⇒ sinA+sinB= -sin A sin B⁡sinC
⇒ sinAsinC+sinB⁡sinC = - SinBsin A 

LHS
 = (sin A + sin B + sin C)²
 = sin² A + sin² B + Sin² c + 2 (sin A sin B + sin B sin C + sin A sin C)
 = 1 - cos² A + 1 - cos² B + 1 - cos² C + 2 (sin A sin B - sin A sin B)
 = 3 - (cos² A + cos² B + cos² C) + 2 × 0
 = 3 - (cos² A + cos² B + cos² C)
 = RHS.

Proved. 

46. Find the value of x in the given equation:
x⁡cos(90° + α)⁡cos(90° - α)+⁡tan180° - α⁡cot(90° + α)=⁡sinα⁡sin(180° -α )

Proof: Given, equation

x⁡cos(90° + α)⁡cos(90° - α)+⁡tan(180° - α)⁡cot(90° + α)=⁡sinα⁡sin(180° - α)
⇒   x (-sinα).sinα+ tanα(-tanα)=sinα. sin α
⇒   -x sin² α + tan² α = sin² α
⇒   tan² α - sin² α = x sin² α
⇒   tan 2 α – sin 2 αsin 2 α = x
⇒   tan 2 αsin 2 α - sin 2 αsin 2 α = x
⇒   sec 2 α - 1 = x
⇒   tan 2 α = x
∴ x = tan 2 α is the required solution.

47. 

Find the value of x in the given equation:
sin(270° - θ)sin(450° - θ)-⁡cos(630° - θ)⁡cos(90° + θ)+ xcos 900 ° = 0

Solution: Do yourself. If any problem, ask me.

48. Find the value of x in the given equation:
xsec(90° + α)cosec α +⁡tan(90° + α)⁡cot(180° - α)= x⁡cotα⁡tan(90° + α)

Solution: Given equation is

x⁡sec(90° + α)cosec α +⁡tan(90°+ α)⁡cot(180° - α)= xcotα⁡tan(90° +α)
⇒  x (-cosec α)cosec α + (-cotα). (-cotα)=xcotα(-cotα)
⇒   - x cosec² α + cot² α = -xcot²α
⇒   - x cosec² α + x cot² α = -cot² α
⇒   - x (cosec² α - cot² α) = -cot² α
⇒   x × 1 = cot² α
∴ x = cot² α is the required solution.

49. Prove that :tan205° -tan115°⁡tan245°+tan335° = 1 + p 21 – p2, if⁡tan25° = P.Proof: Given tan 25° = p Þ  cot⁡cot25° = 1p
Now,
LHS = ⁡tan205° - ⁡tan115°tan245° +  tan335°
= ⁡tan 180° + 25°-tan 90°+25° tan 270°-25°+tan 360°-25°⁡ 
= tan25° + cot 25°⁡ cot25°- tan25° 
= p + 1p1p – p
= P 2 + 1p1p – p1 –p 2p
1 + p 21 – p 2 = RHS proved.
50. If θ = 45°, then prove that:sin3θ- 3sinθ – 4 sin 3θ2 = 12⁡sin3θSolution: Given,
θ = 45°
LHS =sin(3× 45°) - 3⁡sin45° -4 sin 345°2
=⁡sin135° - 3× 1√2 - 4 × ( 1√2) 32
=sin(180° - 45°)-  √3/2 – 2√22
= sin45° - 3 – 22√2
= 12 – 12√2
= 2 – 12√2
= 12√2
RHS = 12sin3θ
= 12sin135°
= 12sin(180° - 45°)
= 12sin45°
= 12 ×1√2
= 12√2
= LHS

 Proved.

51. If A = 2B = 3C =9D = 90°, find the value of cos² A - cot² B + cosec²C + 4 sin² 3D.

Solution: Given. 

A = 2B = 3C = 99 = 90°
2B = 90°          Þ         B =  45°
3C = 90°          Þ         C = 30°
and  9D = 90°  Þ         D = 10 ° 
Now ,
cos²A - cot²B + cos²C + 4 sin² 3D
= cos²90° - cos²45° + cosec²45° + 4 sin² 30°
= 1² - 1²+ √2² + 4 × 12²
= 1 – 1 + 2 + 4 × 14
= 2 + 1
= 3

52. ABCD is a quadrilateral with opposite angles is supplementary then proves that:
(a) cos A + cos B + cos C + cos D = 0(b) tan A + tan B+ tan C + tan D = 0Solution: According to the question,
A + C = 180°  or, A = 180° - C
and B + D = 180  or, B = 180°  - D
    (a) LHS =  Cos A + Cos B + Cos C + Cos D 
       = cos (180° - C) + cos (180° - D) + cos C + cos D 
             = - cos C - cos D + cos C + cos D
             = 0 
= RHS

(b) tan A + tan B + tan C + tan D
= tan (180° - C) + tan (180° - D) + tan C + tan D
=  - tan C - tan D + tan C + tan D
= 0 = RHS

proved.

 53. Prove:
cotx c20⁡cot3x c20⁡cot5x c20⁡cot 7x c20 cot 9x c20 = 1Proof: LHS =⁡cotπ20. ⁡cot3π20⁡cot5π20⁡cot7π20.⁡cot9π20
=cotπ20.⁡cot3π20⁡cotπ4  cotπ2 - 3π2.⁡cotπ2 - π20
=cotπ20.⁡cot3π20×1 ×⁡tan3π20 tanπ20
= 1tan π20 ×1tan 3π20 × tan3π20× tanπ20
= 1 RHS

 Proved.

54. Prove :
cotx c10  cot3x c10⁡cotx c4⁡cot4x c10  cot 2x c10= 1Solution: 

LHS = cotπ10.cot3π10⁡cotπ4⁡cot4π10⁡cot2π10
=⁡cotπ10. ⁡cotπ2-2π10 × 1.⁡cotπ2–π10.⁡cot2π10
=⁡cotπ10.⁡tan2π10 .⁡tanπ10 ×⁡cot2π10
= 1tan π10 × tan2π10× tanπ10 × 1tan 2π10
= 1 

= RHS 

Proved.