Month: June 2018

Bangalore

CBSE_class X time bound sample paper 4 marks Trigonometry_Solutions

Here are the solutions for your reference. In case of any queries  do drop your questions so that it can be answered in time.

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

Bangalore

CBSE_class X time bound sample paper 4 marks Trigonometry

Students must ensure that below questions must be attempted in the limited time of 40 mins.

Else Answer sheet will not be corrected.  This is the time bound test to ensure your understanding

about the chapter and thorough revision.

 

Q 1. If 3x = Cosec A and 3/x = Cot A, find the value of  3[x^2 - 1/x^2] .

Q 2. If  6x = Sec B and 6/x = tan B , find the value of 9[x^2 - 1/ x^2] .

Q 3. If x.sin^3 A + y. cos^3 A = Sin A. Cos A and x.sin A = y.cos A show that

x^2 + y^2 = 1 .

Q 4. If cos A / Cos B = m , Cos A / Sin B =  n show that

(m^2 + n^2 ) cos^2 B = n^2 

Q 5. If tan A + Sin A = m  and tan A – sin A = n prove that

(m^2 -n^2)^2 = 16 mn 

Q 6. If cosec A + cot A = p , prove that (p^2 + 1)cos A = p^2 - 1 

Q 7. Find the value of the other trigonometric ratios if cos A = 2 m/ (m^2 +1)

Q 8. If tan A = p/q , show that

 (p sin A - q cos A) / (p sin A + q cos A) =  (p^2 - q^2 )/ (p^2 + q ^2) 

Q 9. If tan B = a/b , find the value of

( a Sin B + b cos B)/ (a sin B – b cos B)

Q 10. if x = a. cos^3 A , y = b. sin ^3 A  

Prove that  [ x / a ]^2/3 + [ y/b ]^2/3 = 1 

 

Note : Ensure methods to be written clearly. Marks given for method shown also.