# 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) $= 1$