CBSE_class X time bound sample paper 4 marks Trigonometry_Solutions

June 11, 2018 by Ranjit

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)

$= 1$

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