# 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)

$= 1$