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Introduction to Trigonometry Ex- 8.3

 Hi Friends and Champs,

Before proceeding to next exercise, let's discuss about some trigonometric identities. 

Let ΔABC is a right-angle triangle, right angle at B as given below.

Since Δ ABC is a right angle , therefore
CA2 = AB2 + BC2Since]
divide by CA2 at both sides , we have
CA2/CA2 = AB2/CA2 + BC2/CA2
=>(AB/CA)2 + (BC/CA)2= 1
=>cos2A + sin2A =1   -------(1)  ( since cosA = Base / hypotenuse)
                                                       and sinA = perpendicular/ hypotenuse)

divide eq (1) by Sin2A , we have
  [Cos2A /Sin2A]+ [Sin2A/Sin2A] =[1/Sin2A]
=> cot2A + 1 = cosec2A -------(2)

divide eq (1) by Cos2A , we have
  [Cos2A /Cos2A]+ [Sin2A/Cos2A] =[1/Cos2A]
=> 1+ tan2A  = sec2A ---------(3)

Please read carefully the above procedure and do it yourself, there is no need to remember any formula in mathematics, all mathematical equations have a proof, just we need to understand that. Now come to the next exercise 8.3.

                                                                        Exercise 8.3  

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. 
Solution:
Since cot2A + 1 = cosec2A
therefore 1 / sin2A = 1 + cot2A           , (since cosecA = 1/sinA)  
=> sin2A = 1 / (1 + cot2A)
=> sinA = 1/ √(1 + cot2A)  Ans.

since 1+ tan2A  = sec2A
therefore secA = √(1+ tan2A)
=> secA = √(1+ 1/cot2A)                (since tanA = 1 / cotA )
=>secA = cotA / √(1+ cot2A)  Ans.

tanA = 1/cotA    Ans.

2. Write all the other trigonometric ratios of  < A  in terms of sec A

sinA = √ ( 1 - cos2A)             ( since cos2A + sin2A =1 )
       = √ ( 1 - 1/sec2A)         ( since cosA = 1/secA )
        =secA / √ (sec2A - 1)

cosA = 1/secA

tanA = √ (sec2A - 1)

cosecA = 1/sinA  = √ (sec2A - 1) / sec A  
                        ( since sinA =secA / √ (sec2A - 1), already proved )
cotA = 1 / tanA = 1/ √ (sec2A - 1)
 
3. Choose the correct option. Justify your choice. 
 (i) 9 sec2 A – 9 tan2 A = 
 (A) 1      (B) 9      (C) 8      (D) 0
Solution:
9sec2A - 9tan2A = 9/cos2A  - 9sin2A/cos2A
=9 ( 1 - sin2A ) / cos2A
=9 cos2A / cos2A
=9 option (B) is correct  Ans

 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) = 
 (A) 0      (B) 1     (C) 2     (D)–1
Solution:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) =( 1 + sinθ/cosθ + 1/cosθ)(1 + cosθ/sinθ - 1/sinθ)
=(cosθ + sinθ + 1)(sinθ + cosθ -1) / cosθsinθ
=[(cosθ + sinθ)2 - 1] / cosθsinθ  
                            ( since (a + b)(a-b) = a2 -b2)
=[cos2θ + sin2θ + 2cosθsinθ  - 1] / cosθsinθ
                            (since (a + b)2 = a2 + 2ab +b2 )
= [1 + 2cosθsinθ  - 1] / cosθsinθ
= 2cosθsinθ / cosθsinθ
=2   so option (C) is correct Ans

 (iii) (sec A + tan A) (1 – sin A) = 
 (A) sec A      (B)sin A     (C) cosec A    (D) cos A
Solution:
(sec A + tan A) (1 – sin A) =( 1/cosA + sinA/cosA )(1 - sinA )
=( 1 + sinA )(1-sinA) / cosA
=(1 - sin2A) / cosA
= cos2A / cosA
=cosA  option (D) is correct       Ans.

 (iv) 1 + tan 2A / 1 + cot2A = 
(A) sec2A     (B)–1    (C) cot2A     (D) tan2A
Solution:
1 + tan 2A / 1 + cot2A = [ 1 + sin2A /cos2A ] / [ 1 + cos2A/ sin2A ]
=[ (cos2A + sin2A) / cos2A ] / [ (sin2A + cos2A) / sin2A ]
=sin2A/cos2A
=tan2A    option (D) is correct Ans.

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
 (i) (cosec θ – cot θ)2 = (1 - cosθ) / (1 + cosθ)
LHS = (cosec θ – cot θ)2
= (1/sin θ – cosθ/sinθ )2
= (1 - cosθ)2 / sin2θ
=(1 - cosθ)2 / (1 - cos2θ)                       (since cos2θ + sin2θ =1)
=(1 - cosθ)2 / (1 + cosθ)(1 - cosθ)          (since (a + b)(a-b) = a2 -b2)
=(1 - cosθ) / (1 + cosθ)  = RHS         Hence proved.

(ii) cos A / (1 + sin A)  + (1+sinA) /cosA = 2secA 
LHS = [ cos2A  +  (1 + sin A)2 ] /  (1+sinA) /cosA
        = [cos2A + 1 +sin2A + 2sinA ] / (1 + sin A) cosA
        = [1 + 1 + 2sinA ] / (1 + sin A) cosA
        = 2[1 + sinA ] / (1 + sin A) cosA
        = 2 / cosA
        = 2 secA = RHS    Hence proved
 
(iii) tan θ / (1 - cot θ)  + cot θ /( 1 - tan θ) = 1 + sec θ cosec θ
LHS = tan θ / (1 - cot θ)  + cot θ /( 1 - tan θ)
        = [ ( sin θ / cos θ) / ( 1 - cos θ/sin θ)] + [(cos θ/sinθ) / ( 1 - sinθ/cosθ]
        = [ ( sin θ / cos θ) / {( sin θ - cos θ) /sin θ)}] + [(cos θ/sinθ) / {( cosθ - sinθ)} /cosθ]
        = [ ( sin2θ / cos θ( sin θ - cos θ)] + [(cos2θ/sinθ( cosθ - sinθ)]
        =[ ( sin2θ / cos θ( sin θ - cos θ)] - [(cos2θ/sinθ( sinθ - cosθ)]
        =[ ( sin3θ - cos3θ) / cos θ sin θ( sin θ - cos θ)]
        =[( sinθ - cosθ)(( sin2θ + cos2θ + cos θ sin θ)] / cos θ sin θ( sin θ - cos θ)
                                            since  a3 - b3 = (a - b)(a2 + b2 + ab)
        =( sin2θ + cos2θ + cos θ sin θ) / cos θ sin θ
        = ( 1 + cos θ sin θ ) / cos θ sin θ
        =sec θ cosec θ + 1   = RHS     Hence Proved

(iv) (1 + secA) / sec A = sin2 A / (1 - cosA)
LHS =  (1 + secA) / sec A = cos A + 1
RHS = sin2 A / (1 - cosA)
         = (1 - cos2 A ) / (1 - cosA)
        =(1 + cosA)(1 - cosA) / (1 - cosA)
        = (1 + cosA)  
Hence LHS = RHS             Hence Proved

(v) (cosA - sinA +1)/(cosA+sinA-1) = cosec A + cot A
LHS = [(cosA + (1- sinA )] / [(cosA - (1 - sinA)] 
                                    multiply and divide by [(cosA + (1- sinA )]
       = [(cosA  + (1- sinA)]2 / [cos2A  - (1 - sinA)2]
       = [(cos2A  + (1 - sinA)2 + 2cosA(1-sinA)] / [cos2A  - 1 - sin2A + 2sinA]
      = [(cos2A  + 1 + sin2A -2sinA + 2cosA-2cosAsinA)] / [cos2A  - 1 - sin2A + 2sinA)]
     = [(1+ 1  -2sinA + 2cosA-2cosAsinA)] / [cos2A  - cos2A - sin2A - sin2A + 2sinA)]
                                since cos2A + sin2A =1
    =[2(1-sinA) + 2cosA(1 - sinA)] / [2sinA -2sin2A]
    =2(1-sinA)(1 + cosA) / 2sinA(1-sinA)
    =(1 + cosA) / sinA
    =(1/sinA + cosA/sinA)
    =CosecA + cotA = RHS        Hence Proved 

 (vi) √[(1+ sinA) / (1 - sinA) ] = sec A + tan A
LHS = √[(1+ sinA) / (1 - sinA) ]
        =√[(1+ sinA)(1+sinA) / (1 - sinA)(1-sinA) ]
                            multiply and divide by ( 1-sinA)
       =√[(1+ sinA)2 / (1 - sin2A) ]
       =√[(1+ sinA)2 / cos2A ]   since cos2A + sin2A =1
       =(1+ sinA) / cosA ]
       =SecA+ tanA
       = RHS                Hence Proved
(vii) [ (sinθ - 2sin3θ) / ( 2cos3θ - cosθ)] = tanθ
LHS =[ (sinθ - 2sin3θ) / ( 2cos3θ - cosθ)] 
    =[ sinθ(1 - 2sin2θ) / cosθ( 2cos2θ - 1)]
    =[ sinθ(cos2θ + sin2θ - 2sin2θ) / cosθ( 2cos2θ - cos2θ - sin2θ)]
                            since cos2θ + sin2θ = 1
    =[ sinθ(cos2θ - sin2θ ) / cosθ( cos2θ  - sin2θ)]
    =sinθ/cosθ
    =tanθ =RHS 
    Hence Poved
    (viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A
LHS =(sin A + cosec A)2 + (cos A + sec A)2
    = (sinA + 1/sinA)2+ (cos A + 1/cosA)2
    = {(sin2A + 1) / sinA)}2+ {(cos2A + 1) /cosA)}2
    = {(sin4A + 1 + 2sin2A) / sin2A)}+ {(cos4A + 1 + 2cos2A) /cos2A}
    = (sin4Acos2A + cos2A + 2sin2Acos2A + cos4Asin2A + sin2A + 2sin2Acos2A) / cos2Asin2A
    = [sin2Acos2A (cos2A+ sin2A) +cos2A + sin2A +4sin2Acos2A)] / cos2Asin2A
    = [sin2Acos2A + 1 +4sin2Acos2A)] / cos2Asin2A
                                        since cos2A+ sin2A = 1
    = [5sin2Acos2A + 1 )] / cos2Asin2A
    =[5sin2Acos2A / cos2Asin2A+ 1 / cos2Asin2A)] 
    =(5 + cosec2Asec2A )
    =( 5 + ( cot2A + 1)(1 + tan2A)
                                            since cosec2A = ( cot2A + 1) and sec2A= (1 + tan2A)
    =[( 5 + ( cot2A  + cot2A tan2A + 1 + tan2A)]
    =[( 5 + cot2A  + 1 + 1 + tan2A)]
    = 7 + cot2A + tan2A
    = RHS    Hence Proved

 (ix) (cosec A – sin A)(sec A – cos A) = 1 / (tanA + cot A)
LHS=(cosec A – sin A)(sec A – cos A)
    = ( 1 - sin2A)(1 - cos2A) / sinAcosA
    =(1 - cos2A - sin2A + sin2Acos2A) / sinAcosA
    =(sin2A - sin2A + sin2Acos2A) / sinAcosA
                        since 1 - cos2A = sin2A
     = sinAcosA
RHS =  1 / (tanA + cot A)
    = 1 / (sinA/cosA + cos A/sinA)
    =sinAcosA / (sin2A + cos2A)
    =sinAcosA 
LHS = RHS     Hence proved

(x) ( 1 + tan2A) / ( 1 + cot2A)  = [(1-tanA)/(1-cotA)]2 = tan2A
LHS =( 1 + sin2A/cos2A) / ( 1 + cos2A/sin2A) 
        =( cos2A + sin2A)sin2A / (sin2A + cos2A)cos2A  
        = sin2A / cos2A = tan2A    
                                since sin2A + cos2A = 1
        Hence proved
[(1-tanA)/(1-cotA)]2 = [ ( 1 - sinA/cosA) / ( 1 - cosA / sinA)]2
                                 = [ ( cosA - sinA )sinA/ ( sinA - cosA )cosA]2
                                 = [ ( cosA - sinA )sinA/ - ( cosA - sinA )cosA]2
                                 = [ - sinA/cosA]2
                                 = [ - tanA]2
                                =  tan2A
        Hence Proved


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