## ICSE sample papers for class 10 computer application (…

##### Chemistry and Computer Applications

Practice time bound sample paper Chapter wise :

CBSE_Trigonometry_classX_3_marks_questions.docx

## CBSE Class 8th Maths Linear equations

All the students who have finished the chapter of linear equations are required to take this test and submit the workings. Your marks will be displayed after finishing the test.

Revision Test for CBSE Class 8th Maths chapter Linear Equations.

## ICSE Class 7th Maths Decimal test

Revision Exercise for ICSE Class 7th Maths chapter Decimal.

Duration of this test is 30 mins.

Your score will be visible at the end of test.

All your workings are needed to be evaluated in the next class.

All the best

Revision Test for the Algebraic expression and Linear equation in one variable. You can download the question paper and start attempting each question. Duration of Test is 30 minutes. All questions are compulsory.

## CBSE Class 10th revision test

Revision test for class 10th CBSE

Chapters: Triangles, Trigonometry and Height and Distance.

Duration : 40 minutes

## ICSE class 9th revision test

Chapters: Rational and Irrational number , Compound interest, Expansion ,  Factorisation , Simultaneous linear equations , Indices , Logarithms

Time : 1 hour

## ICSE class 6th physics revision test

Chapter : Measurement

Marks : 20 marks

Time : 40 mins

## ICSE class 10th JAVA questions

Duration : 20 min

Marks : 20

## java test

JAVA Revision Test.
45 mins
30 question

## IGCSE Class 7th Maths

IGCSE  Class 7 Maths Revision Test

Test includes chapters Fractions, decimals and percentages
Sequences functions and graphs and Angle properties.
Duration : 45 mins

Revision Test for the Class 8th CBSE Maths. You can download the question paper and start attempting each question. Duration of Test is 60 minutes. All questions are compulsory.

## 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)

$= 1$

## 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.

## Java X ICSE Solutions

Q 1. Write a program to print following pattern using BLUE J IDE.

1234321
123*321
12***21
1*****1

Ans :

Following program is tried and tested in Blue J 3.08 version.

class jaiii
{
public static void main(String args[])

{
int i,j,l;

for(i=4;i>=1;i–)
{

for(j=1;j<=i;j++)
System.out.print(j);
for(int m=i;m<4;m++) //increment o *
System.out.print(“*”);

for(int n=3;n>i;n–) //decrement of *
System.out.print(“*”);
for(l=i;l>=1;l–)
{ if(l==4)
continue;
System.out.print(l);
}
System.out.println();
}

Q 2.  WAP to print following pattern in java

A
A B
A B C
A B C D
A B C D E

Ans : Given program below is test on Blue J 3.08 version .

public class sid6

{

public static void main(String args[])

{

char ch;

for(int a=1;a<=5;a++)

{

ch=’A’;

for(int c=4;c>=a-1;c–)

{

System.out.print(” “);

}

for(int b=1;b<=a;b++)

{

System.out.print(ch+” “);

ch++;

}

System.out.println();

}

}

}

Now its the time to buy the 10 year question paper for ICSE examination.  Almost all the portions

are over for 10th graders. I suggest now every student of ICSE board to practice questions from the old sample papers and do the time bound tests. You will definitely improve your efficiency in coming days. Any questions or

doubts you can reach to me.