Class 9: Trigonometry (Compound Angles)

The proved trigonometric statements are also benificial to the students of 10th standard also...


1. Prove that: 
Tan(A + B) = TanA+TanB1-TanA TanB . 
Proof: LHS = Tan(A + B)
         = Sin(A + B)Cos(A + B)  
       = SinACosB+CosASinBCosACosB - SinASinB
        =  SinACosBCosACosB+ CosASinBCosACosBCosACosBCosACosB-SinASinBCosACosB   [ ∵ Dividing Numerator & denominator by CosACosB & seperating LCM.]
      =TanA+TanB1-TanA TanB
       = RHS.
                    Proved.

2. Prove that : 

Tan(A - B) =TanA-TanB1+ TanA TanB

Proof: Here, 
LHS  = Tan(A - B) 
        =Sin(A - B)Cos(A - B)
        =SinACosB - CosASinBCosACosB + SinASinB 
        = SinACosBCosACosB- CosASinBCosACosBCosACosBCosACosB+SinASinBCosACosB   [ Dividing numerator & denominator by CosACosB.]
       =TanA-TanB1+TanA TanB
        = RHS.

Proved.

3. Prove that: 

Cot(A + B)  =CotACotB-1CotB+CotA .

Proof: LHS = Cot(A + B)
        = Cos(A+B)Sin(A+B)
       =CosACosB- SinASinBSinACosB + CosASinB 
       =   CosACosB SinASinB- SinASinBSinASinBSinACosBSinASinB+CosASinBSinASinB  [∵ Dividing numerator and denominator by SinASinB & Seperating LCM.]
      = CotACotB-1CotB+CotA
       = RHS.

Proved.

Alternatively, you can also prove it by the formula of tan(A+B).

4. Prove that:

 Cot(A - B)  =CotACotB+1CotB-CotA .

Proof: LHS = Cot(A-B)
       =Cos(A-B)Sin(A-B) 
      =CosACosB+SinASinBSinACosB - CosASinB
     = CosACosB SinASinB+ SinASinBSinASinBSinACosBSinASinB-CosASinBSinASinB [ ∵ Dividing numerator & denominator by SinASinB & seperating LCM.]
    =CotACotB+1CotB-CotA
     = RHS.

Proved.

Alternatively, you can also prove it by the formula of tan(A-B).

5. Prove that: 

Sin(A+B).Sin(A-B) = Cos²B - Cos²A.

Proof: LHS = Sin(A+B).Sin(A-B)
         = ( SinACosB + CosASinB)(SinACosB - CosASinB)
         = (SinACosB)² - (CosASinB)²
         = Sin²ACos²B - Cos²ASin²B
         = ( 1- Cos²A)Cos²B - Cos²A(1 - Cos²B)
         = Cos²B - Cos²ACos²B - Cos²A + Cos²ACos²B
         =  Cos²B - Cos²A.
         = RHS.

Proved.

6. Prove that : 

Sin(A+B).Sin(A-B) = Sin²A - Sin²B.

Proof: Here,
LHS = sin ( A + B ) ⋅ sin ( A − B )
        = ( sinA cosB + cosA sinB ) ( sinA cosB − cosA sinB )
        = ( sinA cosB )² − ( cosA sinB )²    [Using the identity ( p + q ) ( p − q ) = p ² − q ² .]
        = sin²A cos²B − sin²B cos²A
        = sin²A ( 1 − sin²B ) − sin²B ( 1 − sin²A )
        = sin²A − sin B − sin²A sin²B + sin²B sin²A
        = sin²A − sin²B.
        = RHS.

Proved.

7. Prove that : 

cos(A+B).cos(A-B) = Cos²A - Sin²B.

Proof: Here,
LHS =  cos(A+B).cos(A-B)
        = (cosAcosB - sinAsinB)(cosAcosB +sinAsinB)
        = (cosAcosB)² - (sinAsinB)²
        = cos²Acos²B - sin²Asin²B
        = Cos²A( 1 - Sin²B) - (1 - cos²A)Sin²B
        = Cos²A - Cos²ASin²B - Sin²B + cos²ASin²B
        = Cos²A - Sin²B.
        = RHS 

Proved.

8. Prove that: 

cos(A+B).cos(A-B) = cos²B - sin²A.

Proof: Here,
LHS =  cos(A+B).cos(A-B)
        = (cosAcosB - sinAsinB)(cosAcosB +sinAsinB)
        = (cosAcosB)² - (sinAsinB)²
        = cos²Acos²B - sin²Asin²B
        = cos²B( 1 - cin²A) - (1 - cos²B)cin²A
        = cos²B - cos²Bsin²A - sin²A + cos²Bsin²A
        = cos²B - sin²A.
        = RHS.

Proved.

9. Prove that: 

cot(A+B).cot(A-B)  =Cot²ACot²B-1Cot²B-Cot²A .

Proof: LHS = cot(A+B).cot(A-B)
        = CotACotB-1CotB+CotA⨉CotACotB+1CotB-CotA
       = (CotACotB-1)(CotACotB +1)(CotB+CotA)(CotB - CotA)
       =Cot²ACot²B-1Cot²B-Cot²A
        = RHS.

Proved.

10. Prove that: 

tan(A+B).tan(A-B) = tan²A-tan²B1-tan²Atan²B   .

Proof: LHS = tan(A+B).tan(A-B)
                 =tanA+tanB1-tanAtanB⨉tanA-tanB1+tanAtanB
                 =tan²A-tan²B1²-(tanAtanB)²
                 =tan²A-tan²B1-tan²Atan²B  
                 = RHS.

Proved.

11. Find the value of cos15^o without using calculator or table.
Solution: We know,
 Cos15^o = cos(45^o - 30^o)
              = cos45^ocos30^o + sin45^osin30^o
              = 1√2⨉ 
√32  + 12 ⨉1√2 
            
 = √3+12√2
Which is required value of cos15^o.

12. If A+B = π^c/4, prove that: 

cotAcotB - cotA - cotB = 1.

Proof: We know, π^c = 180^o
So, A+B = 45^o
Operating both sides by cot we get
      cot(A+B) = cot45^o    , using formula of cot(A+B), we get
⇒ cotAcotB-1cotB+cotA = 1          [ Since, cot45^o = 1.]
⇒ cotAcotB - 1 = cotB + cotA
⇒ cotAcotB - cotA - cotB = 1
       LHS = RHS.

Proved.

Next method :
LHS = cotAcotB - cotA - cotB
        = 1 + [cotAcotB - 1] - cotA - cotB
   = 1 + [cotB + cotA] × cotAcotB-1cotB+cotA  - cotA - cotB
   = 1 + [cotB + cotA] × cot(A+B) - cotA - cotB
   = 1 + [cotB + cotA] × 1 - cotA - cotB 

 [cot(A+B) = cot(π^c/4) = 1]

   = 1 + cotB + cotA] - cotA - cotB
   = 1 

Proved

13. Prove that: 
sin105° + cos105° = 1√2  
Proof: Here,
 LHS = sin105° + cos105°
         = sin(60^o + 45^o ) + cos ( 60^o + 45^o )
         = ( sin 60^o cos 45^o + cos 60^o sin 45^o ) + ( cos 60^o cos 45^o - sin60^o sin 45^o)
         =  √32 . 1√2 + 12 . 1√2 +  12 . 1√2 -  √32 . 1√2 


 
         = 22√2 
       = 1√2  
       = RHS.

Proved.

14. Prove that:  cot73^o.cot28^o+1cot73^o-cot28^o  = 1
Proof: LHS = cot73^o.cot28^o+1cot73^o-cot28^o  
    = cot(73^o -28^o) [ Since, cotA.cotB+1cotB-cotA = cot(A-B).]
    = cot45^o
    = 1
     = RHS.

Proved.

15. Prove that: 

cot27^o.cot18^o - cot27^o- cot18^o = 1

Proof: Here,
We know,
      27^o + 18^o = 45^o
Operating both side by cot,we get
      Cot(27^o+18^o) = cot45^o
⇒ cot27^o.cot18^o-1cot27^o+cot18^o = 1       
[ Since, using formula of cot(A+B) & cot45^o = 1.]
⇒ cot27^o.cot18^o - 1 = cot27^o + cot18^o
⇒ cot27^o.cot18^o- cot27^o - cot18^o = 1
i.e.  LHS = RHS

Proved.

Next method :
LHS = cot27°cot18º - cot27º - cot18º
        = 1 + [cot27ºcot18º - 1] - cot27º - cot18º
   = 1 + [cot27º + cot18º] × cot27ºcot18º-1cot27º+cot18º  - cot27º - cot18º
    = 1 + [cot27º + cot18º] × cot(18º+27º) - cot27º - cot18º
   = 1 + [cot27º + cot18º] × 1 - cot27º - cot18º 
   = 1 + cot27º + cot18º - cot27º - cot18º
   = 1 

Ptoved

16. Prove that: tan50^o + tan60^o + tan70^o = tan50^o.tan60^o.tan70^o.

Proof: Here,
     50^o + 60^o + 70^o = 180^o
⇒ 50^o + 60^o = 180^o - 70^o
Operating both sides by tan, we get
     tan(50^o + 60^o) = tan(180^o - 70^o)
⇒  tan50^o +tan60^o1 - tan50^o.tan60^o = - tan70^o        [ Since, using formula of tan(A+B) & tan(180^o- A) = -tanA.]
⇒ tan50^o + tan60^o = -tan70^o(1 - tan50^o.tan60^o)
⇒ tan50o + tan60^o = -tan70^o + tan50^o.tan60^otan70^o 
⇒ tan50^o + tan60^o + tan70^o = tan50^o.tan60^otan70^o 
∴  LHS = RHS

Proved.

17. Prove that: tan9A - tan6A - tan3A = tan9A.tan6A.tan3A.
Proof: Here,
     3A + 6A = 9A
Operating both sides by tan, we get
     tan( 3A + 6A) = tan(9A)
⇒ tan3A + tan6A1 - tan3A.tan6A = tan9A  [ Since, using formula of tan(A+B).]
⇒ tan3A + tan6A = tan9A(1 - tan3A.tan6A)
⇒ tan3A + tan6A = tan9A - tan3A.tan6A.tan9A 
⇒ tan9A - tan6A - tan3A = tan9Atan6Atan3A.
∴  LHS = RHS

Proved.

18. Prove that : sin(x+y) + sin(x-y) = 2sinxcosy.
Proof: Here,
LHS = sin(x+y) + sin(x-y)
         = sinxcosy + cosxsiny + sinxsiny - cosxcosy
         = sinxcosy + sinxcosy
         = 2sinxcosy
         = RHS.

Proved.

19. If A+B+C = π^c & cosA = cosBcosC, prove that tanA =  tanB + tanC.
Proof: LHS = tanA 
        = sinAcosA
       =sin[π -(B+C)])cosBcosC       [∵ A= π - (B+C) & cosA = cosBcosC.]
       = sin(B+C)cosBcosC
       = sinBcosC + cosBsinCcosBcosC
       = sinBcosCcosBcosC +  cosBsinCcosBcosC
       =sinBcosB +  sinCcosC 
       = tanB + tanC
       = RHS.

Proved.

20. An angle θ is divided into two parts α & β such that  tanα : tanβ = x : y, then prove that:
sin(α-β)  = x-yx+ysinθ      
Proof:   Here, we have given
     tanα : tanβ = x : y.
⇒  tanαtanβ =  xy 
By using componendo & dividendo , we have
       tanα + tanβtanα - tanβ = x+yx - y
Converting into sin & cos we get
     sinαcosβ + cosαsinβsinαcosβ - cosαsinβ = x + yx - y
⇒   sin(α+β)sin(α - β) = x+yx - y
⇒  sinθsin(α-β) = x+yx - y               [ Since, α+β = θ.]
⇒ sin(α-β)  = x-yx+ysinθ          

Proved.                       

21. Prove that:  
sinA + sin(A+2π3) + sin(A+ 4π3) = 0
Proof: Here,
LHS = sinA + sin(A+2π3) + sin(A+ 4π3)     
Now, put π = 180^o, we get
        = sinA + sin(A+ 120^o ) + sin(A+  240^o )     
        = sinA + sinAcos120^o + cosAsin120^o + sinAcos240^o+cosAsin120^o
        = sinA + sinA(- 12) + cosA(√32) + sinA(- 12) + cosA(-√32)
        = sinA - 12 sinA - 12 sinA
        = sinA - sinA
        = 0
       = RHS

Proved.

22. sin²(π6 + θ) - sin²(π6 - θ)  = √32 sin2θ 
Proof: Here, 
LHS = sin²(π6 + θ) - sin²(π6 - θ)  
        = sin²( 30^o + θ) - sin²( 30^o - θ)                 
[ Since, π = 180^o.]
        = sin(30^o+θ + 30^o-θ) . sin(30^o+θ-30^o+θ)       
 [ Since, sin²A - sin²B = sin(A+B).sin(A-B).]
        = sin60^o .sin2θ
        = √32 sin2θ 
        = RHS

Proved.

23. If tanA = 56 & tanB=2π3, prove that prove that,
A+B =  π^c4 
Proof: We know,
 tan(A+B) = tanA+tanB1 - tanAtanB
                 = 56+1111- 56×111
                 =56665666
                 = 1
                 = tan45^o
 tan(A+B) = tan π4 
  So, A+B =  π^c4 

Proved.

24. 

In any quadrilateral ABCD, prove that  

cosAcosB - cosCcosD =  sinAsinB - sinCsinD.

Proof: We know,
           ABCD is a quadrilateral.
Then
      A+B+C+D = 360^o 
⇒  A+B = 360 - (C+D)
Operating both sides by cos,
     cos(A+B) = cos{360 -(C+D)}         
Using formula,we get
⇒ cosAcosB - sinAsinB = cosCcosD - sinCsinD    [ Since, cos(360 -θ) = cosθ.]
⇒  cosAcosB - cosCcosD =  sinAsinB - sinCsinD
      LHS = RHS

Proved.