CBSE_class X time bound sample paper 4 marks Trigonometry_Solutions

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Ans 1 .  given 3x = cosecA, 3/x = cotA

  =3[  x^2 - 1/x^2  ] 

=3x^2 - 3/x^2 

=1/3[ 9x^2 - 9/x^2]      (multiply and divide by 3 )

=1/3[ {3x}^2 - {3/x}^2  ]  

  =1/3[ cosec^2A - cot^2A ]      (identity 3)

=1/3 * 1 

=1/3   

Ans 2 .  given 6x = secB , 6/x = tanB

= 9 [ x^2 - 1/x^2 ]

= 9 * 36/36 [ x^2 - 1/x^2 ]     ( multiply and divide by 36)

= 9 /36 [ (6x)^2 - (6/x)^2 ]

= 1/4 [ sec^2B - tan^2B ]       (substituting the values of 6x and 6/x)

= 1/4 [ 1]      ( identity 2 )

= 1/4   

Ans 3 . Given

x.sin^3A + y.cos^3A = sinA.cosA  ..............eq(1)    

             , x.sinA = y.cosA   ..................eq(2)   

  x =  y.cosA / SinA   

substituting in eq(1).

y.cosA/sinA . sin^3A + y.cos^3A = sinA.cosA   

y.cosA.sin^2A + y.cos^3A = sinA.cosA   

y.cosA ( sin^2A + cos^2A) = sinA.cosA   

y.cosA. (1) = sinA.cosA  

y = sinA   

x = sinA. cosA/sinA   

x = cosA   

L.H.S = x^2 + y^2    

= sin^2A + cos^2A   (identity 1)

= 1   

Hence L.H.S = R.H.S

Ans 4.

  LHS = [ cos^2 A/ cos^2B + cos^2 A / sin^2B ] . cos^2B  

 = cos^2A [ 1/cos^2B + 1/sin^2B ]. cos^2B

 = cos^2A[ (sin^2B + cos^2B) / (cos^2B.sin^2B) ]. cos^2B

= cos^2A. 1. cos^2B / cos^2B.sin^2B

= cos^2A / sin^2B

= n^2 

= RHS   

Ans 5.

LHS = ( m^2 - n^2 )^2

= [( tanA + sinA )^2 - ( tanA - sinA )^2 ]^2 

= [ tan^2A + sin^2A + 2.sinA.tanA - ( tan^2A + sin^2A - 2.sinA.tanA ) ]^2  

= [ tan^2A + sin^2A + 2.sinA.tanA -  tan^2A - sin^2A + 2.sinA.tanA ) ]^2  

= [ 4.sinA.tanA ]^2  

= 16.tan^2A.sin^2A 

= 16.tan^2A.(1 - cos^2A )

= 16.[tan^2A - tan^2A. cos^2A ]

= 16.[tan^2A - ( sin^2A / cos^2A ). cos^2A ]

= 16.[tan^2A -  sin^2A )

= 16.( tanA + sinA ).( tanA - sinA )

= 16.m.n

= RHS

Ans 6.

LHS = ( p^2 +1 ).cosA

  = [ ( cosecA + cotA )^2 +1 ].cosA

  = [  cosec^2A + cot^2A  + 2.cosecA.cotA +1 ].cosA

  = [  cosec^2A + cosec^2A  + 2.cosecA.cotA  ].cosA    ( identity 2 )

  = [  2.cosec^2A  + 2.cosecA.cotA  ].cosA

  = [  2.cosecA( cosecA + cotA  ].cosA

  = [  2.cosA/sinA ( cosecA + cotA  ]

  = [  2.cotA ( cosecA + cotA  ]

  = 2.cotA . cosecA + 2.cot^2A   

  = 2.cotA . cosecA + .cot^2A + cot^2A 

  = 2.cotA . cosecA + .cot^2A + cosec^2A - 1

  = ( cosecA + .cotA )^2 - 1

  = p^2 - 1

  = RHS

Ans 7.

  cosA = 2m/(m^2 + 1)

  sin^2A = 1 - cos^2A 

  sin^2A = 1 - [ 2m/(m^2 + 1 ) ]^2 

  sin^2A = [ (m^2 + 1)^2 - 4m^2 ] / ( m^2 + 1 )^2

  sin^2A = [ (m^2 + 1 +2m) ( m^2 + 1 -2m ) ] / ( m^2 + 1 )^2

  sin^2A = [ (m + 1 )^2 ( m - 1 )^2 ] / ( m^2 + 1 )^2

  sinA = [ (m + 1 )^2 ( m - 1 )^2 ] / ( m^2 + 1 )^2

  sinA = ( m^2 - 1 ) / ( m^2 + 1 )

 

  tanA = sinA / cosA 

  tanA = [(m^2 -1) / (m^2 + 1 ) ] / [ 2m / (m^2 + 1) ]

  tanA =  (m^2 - 1) / 2m

  cosecA =  (m^2 + 1) / (m^2 - 1 )

  cotA =  2m / (m^2 - 1)

Ans 8.

given tanA = p/q

LHS = (psinA – qcosA ) / (psinA + qcosA)

dividing both numerator and denominator by qcosA

   = [ p.sinA/q.cosA - 1 ] / [ p.sinA / q.cosA + 1 ]

   = [ p.tanA/q - 1 ] / [ p.tanA/ q + 1 ]

   = [ p^2/q^2 - 1 ] / [ p^2/ q^2 + 1 ]

   = [ (p^2-q^2) / q^2 ] / [ (p^2 +  q^2) / q^2 ]

   =  (p^2-q^2)  /  (p^2 +  q^2)

   =  RHS

hence proved.

Ans 9.

Given tan B = a/b

= (a sinB + bcosB) / (a sinB – bcosB)

dividing both numerator and denominator by bcosB

   =  (a/b .tanB + 1) / (a/b tan B - 1)

   =  (a^2/b^2  + 1) / (a^2/b^2  - 1)

   =  (a^2 + b^2 ) / (a^2  - b^2 )

 

Ans 10.

given x = a Cos³A  y = b Sin³A

  LHS  =  [ x / a ] ^2/3 + [ y / b ] ^ 2/3

  =  [ acos^3A/ a ] ^2/3 + [ bsin^3A / b ] ^ 2/3

  =   cos ^2A + sin ^ 2A      ( identity 1)

  =   1

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