Class 10 Mathematics: CBSE Sample Question Paper with Solutions (2022-23) (Standard) - 1

 Page 1


1 
 
SAMPLE QUESTION PAPER 
MARKING SCHEME 
SUBJECT: MATHEMATICS- STANDARD 
CLASS X 
 
SECTION - A 
1 (c) 35 
 
1 
2 (b) x
2
–(p+1)x +p=0 
 
1 
3 (b) 2/3 
 
1 
4 (d) 2 
 
1 
5 (c) (2,-1) 
                  
1 
6 (d) 2:3 
 
1 
7 (b) tan 30° 
                     
1 
8 (b) 2 
                            
1 
9 
(c) x= 
????
?? +?? 
 
1 
10 (c) 8cm 
 
1 
11 (d) 3v3cm 
 
1 
12 (d) 9p cm
2 
 
1 
13 (c) 96 cm
2
 1 
   
14 (b) 12 
 
1 
15 (d) 7000 
 
1 
16 (b) 25 
 
1 
17 (c) 11/36 
 
1 
18 (a) 1/3 
 
1 
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct 
explanation of assertion (A) 
1 
 
 
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct  
explanation of assertion (A) 
1 
Page 2


1 
 
SAMPLE QUESTION PAPER 
MARKING SCHEME 
SUBJECT: MATHEMATICS- STANDARD 
CLASS X 
 
SECTION - A 
1 (c) 35 
 
1 
2 (b) x
2
–(p+1)x +p=0 
 
1 
3 (b) 2/3 
 
1 
4 (d) 2 
 
1 
5 (c) (2,-1) 
                  
1 
6 (d) 2:3 
 
1 
7 (b) tan 30° 
                     
1 
8 (b) 2 
                            
1 
9 
(c) x= 
????
?? +?? 
 
1 
10 (c) 8cm 
 
1 
11 (d) 3v3cm 
 
1 
12 (d) 9p cm
2 
 
1 
13 (c) 96 cm
2
 1 
   
14 (b) 12 
 
1 
15 (d) 7000 
 
1 
16 (b) 25 
 
1 
17 (c) 11/36 
 
1 
18 (a) 1/3 
 
1 
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct 
explanation of assertion (A) 
1 
 
 
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct  
explanation of assertion (A) 
1 
2 
 
 SECTION – B  
   
21 Adding the two equations and dividing by 10, we get : x+y = 10 
Subtracting  the two equations  and dividing by -2, we get : x-y =1 
Solving these two new equations, we get, x = 11/2 
                                                                   y = 9/2 
½ 
½ 
½ 
½ 
 
   
22 In ?ABC, 
?1 = ?2 
? AB = BD ………………………(i) 
Given, 
AD/AE = AC/BD 
Using equation (i), we get 
AD/AE = AC/AB   ……………….(ii) 
In ?BAE and ?CAD, by equation (ii), 
AC/AB   = AD/AE  
?A= ?A (common) 
? ?BAE ~ ?CAD [By SAS similarity criterion] 
 
 
 
½ 
 
 
½ 
 
 
½ 
 
½ 
 
   
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent) 
?AOB = 105° (By angle sum property of a triangle) 
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the 
angle subtended  by the arc at the centre) 
 
½ 
½ 
1 
 
 
24 
 
We know that, in 60 minutes, the tip of minute hand moves 360° 
In 1 minute, it will move =360°/60  = 6°  
? From  7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210° 
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?         
                                                                                     of  210° and radius  of 6 cm   
                                                                                          = 
210
360
x p x 6
2
 
                                                                                =  
  
7
12
x 
22
7
 x 6 x 6 
                                                                               =66cm
2
 
                                                                            
 
OR 
 
 Let the measure of ?A, ?B, ?C and ?D be  ?1,  ?2, ?3 and  ?4  respectively  
Required area = Area of sector with centre A + Area of sector with centre B  
                          + Area of sector with centre C + Area of sector with centre D 
 
 
½ 
½ 
 
 
 
½ 
 
 
½ 
 
 
 
 
 
 
½ 
 
Page 3


1 
 
SAMPLE QUESTION PAPER 
MARKING SCHEME 
SUBJECT: MATHEMATICS- STANDARD 
CLASS X 
 
SECTION - A 
1 (c) 35 
 
1 
2 (b) x
2
–(p+1)x +p=0 
 
1 
3 (b) 2/3 
 
1 
4 (d) 2 
 
1 
5 (c) (2,-1) 
                  
1 
6 (d) 2:3 
 
1 
7 (b) tan 30° 
                     
1 
8 (b) 2 
                            
1 
9 
(c) x= 
????
?? +?? 
 
1 
10 (c) 8cm 
 
1 
11 (d) 3v3cm 
 
1 
12 (d) 9p cm
2 
 
1 
13 (c) 96 cm
2
 1 
   
14 (b) 12 
 
1 
15 (d) 7000 
 
1 
16 (b) 25 
 
1 
17 (c) 11/36 
 
1 
18 (a) 1/3 
 
1 
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct 
explanation of assertion (A) 
1 
 
 
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct  
explanation of assertion (A) 
1 
2 
 
 SECTION – B  
   
21 Adding the two equations and dividing by 10, we get : x+y = 10 
Subtracting  the two equations  and dividing by -2, we get : x-y =1 
Solving these two new equations, we get, x = 11/2 
                                                                   y = 9/2 
½ 
½ 
½ 
½ 
 
   
22 In ?ABC, 
?1 = ?2 
? AB = BD ………………………(i) 
Given, 
AD/AE = AC/BD 
Using equation (i), we get 
AD/AE = AC/AB   ……………….(ii) 
In ?BAE and ?CAD, by equation (ii), 
AC/AB   = AD/AE  
?A= ?A (common) 
? ?BAE ~ ?CAD [By SAS similarity criterion] 
 
 
 
½ 
 
 
½ 
 
 
½ 
 
½ 
 
   
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent) 
?AOB = 105° (By angle sum property of a triangle) 
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the 
angle subtended  by the arc at the centre) 
 
½ 
½ 
1 
 
 
24 
 
We know that, in 60 minutes, the tip of minute hand moves 360° 
In 1 minute, it will move =360°/60  = 6°  
? From  7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210° 
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?         
                                                                                     of  210° and radius  of 6 cm   
                                                                                          = 
210
360
x p x 6
2
 
                                                                                =  
  
7
12
x 
22
7
 x 6 x 6 
                                                                               =66cm
2
 
                                                                            
 
OR 
 
 Let the measure of ?A, ?B, ?C and ?D be  ?1,  ?2, ?3 and  ?4  respectively  
Required area = Area of sector with centre A + Area of sector with centre B  
                          + Area of sector with centre C + Area of sector with centre D 
 
 
½ 
½ 
 
 
 
½ 
 
 
½ 
 
 
 
 
 
 
½ 
 
3 
 
                         =  
?? 1
360
 x  p x 7
2
 + 
?? 2
360
 x  p x 7
2 
+ 
?? 3 
360
 x  p x 7
2  
+ 
?? 4
360
 x  p x 7
2
 
 
                             = 
(?? 1 + ?? 2 + ?? 3 + ?? 4)
360
  x  p x 7
2 
  
                       = 
(?????? )
360
 x 
????
7
 x 7x 7 ( By angle sum property of a triangle) 
                        = 154 cm
2
 
 
 
½ 
 
 
 
½ 
½ 
 
25 
 
 
 
 
 
 
 
 
sin(A+B) =1 = sin 90, so A+B = 90……………….(i) 
 cos(A-B)= v3/2 = cos 30, so A-B= 30……………(ii) 
From (i) & (ii) ?A = 60° 
                And ?B = 30°  
OR 
 
cos? - sin ?
cos?+sin ?
 = 
1-v3
1+v3
  
Dividing the numerator and denominator of LHS by cos?, we get 
1 - tan ?
1+tan ?
 = 
1-v3
1+v3
 
Which on simplification (or comparison) gives tan? = v3 
 Or ?= 60° 
½ 
½ 
½ 
½ 
 
 
 
 
½ 
½ 
 
½ 
½ 
 
                                                        SECTION - C  
26 
 
Let us assume 5 + 2v3 is rational, then it must be in the form of p/q where p and 
q are co-prime integers and q ? 0   
i.e 5 + 2v3 = p/q 
 So v3 = 
?? -5?? 2?? ……………………(i) 
Since p, q, 5 and 2 are integers and q ? 0, HS of equation (i) is rational. But 
LHS of (i) is v3  which is irrational. This is not possible. 
This contradiction has arisen due to our wrong assumption that 5 + 2v3 is 
rational. So, 5 + 2v3 is irrational. 
 
 
1 
 
 
½ 
 
½ 
 
½ 
 
 
 
½ 
 
 
27 Let a and ß be the zeros of the polynomial 2x
2 
-5x -3 
Then  a + ß = 5/2 
And  aß = -3/2. 
Let 2a and 2ß be the zeros x
2 
+ px +q 
Then  2a + 2ß = -p 
          2(a + ß) = -p 
          2 x 5/2 =-p 
So p = -5 
And  2a x 2ß = q 
            4 aß = q  
   So q = 4 x-3/2 
           = -6 
 
½ 
½ 
 
½ 
 
 
½ 
½ 
 
 
½ 
Page 4


1 
 
SAMPLE QUESTION PAPER 
MARKING SCHEME 
SUBJECT: MATHEMATICS- STANDARD 
CLASS X 
 
SECTION - A 
1 (c) 35 
 
1 
2 (b) x
2
–(p+1)x +p=0 
 
1 
3 (b) 2/3 
 
1 
4 (d) 2 
 
1 
5 (c) (2,-1) 
                  
1 
6 (d) 2:3 
 
1 
7 (b) tan 30° 
                     
1 
8 (b) 2 
                            
1 
9 
(c) x= 
????
?? +?? 
 
1 
10 (c) 8cm 
 
1 
11 (d) 3v3cm 
 
1 
12 (d) 9p cm
2 
 
1 
13 (c) 96 cm
2
 1 
   
14 (b) 12 
 
1 
15 (d) 7000 
 
1 
16 (b) 25 
 
1 
17 (c) 11/36 
 
1 
18 (a) 1/3 
 
1 
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct 
explanation of assertion (A) 
1 
 
 
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct  
explanation of assertion (A) 
1 
2 
 
 SECTION – B  
   
21 Adding the two equations and dividing by 10, we get : x+y = 10 
Subtracting  the two equations  and dividing by -2, we get : x-y =1 
Solving these two new equations, we get, x = 11/2 
                                                                   y = 9/2 
½ 
½ 
½ 
½ 
 
   
22 In ?ABC, 
?1 = ?2 
? AB = BD ………………………(i) 
Given, 
AD/AE = AC/BD 
Using equation (i), we get 
AD/AE = AC/AB   ……………….(ii) 
In ?BAE and ?CAD, by equation (ii), 
AC/AB   = AD/AE  
?A= ?A (common) 
? ?BAE ~ ?CAD [By SAS similarity criterion] 
 
 
 
½ 
 
 
½ 
 
 
½ 
 
½ 
 
   
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent) 
?AOB = 105° (By angle sum property of a triangle) 
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the 
angle subtended  by the arc at the centre) 
 
½ 
½ 
1 
 
 
24 
 
We know that, in 60 minutes, the tip of minute hand moves 360° 
In 1 minute, it will move =360°/60  = 6°  
? From  7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210° 
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?         
                                                                                     of  210° and radius  of 6 cm   
                                                                                          = 
210
360
x p x 6
2
 
                                                                                =  
  
7
12
x 
22
7
 x 6 x 6 
                                                                               =66cm
2
 
                                                                            
 
OR 
 
 Let the measure of ?A, ?B, ?C and ?D be  ?1,  ?2, ?3 and  ?4  respectively  
Required area = Area of sector with centre A + Area of sector with centre B  
                          + Area of sector with centre C + Area of sector with centre D 
 
 
½ 
½ 
 
 
 
½ 
 
 
½ 
 
 
 
 
 
 
½ 
 
3 
 
                         =  
?? 1
360
 x  p x 7
2
 + 
?? 2
360
 x  p x 7
2 
+ 
?? 3 
360
 x  p x 7
2  
+ 
?? 4
360
 x  p x 7
2
 
 
                             = 
(?? 1 + ?? 2 + ?? 3 + ?? 4)
360
  x  p x 7
2 
  
                       = 
(?????? )
360
 x 
????
7
 x 7x 7 ( By angle sum property of a triangle) 
                        = 154 cm
2
 
 
 
½ 
 
 
 
½ 
½ 
 
25 
 
 
 
 
 
 
 
 
sin(A+B) =1 = sin 90, so A+B = 90……………….(i) 
 cos(A-B)= v3/2 = cos 30, so A-B= 30……………(ii) 
From (i) & (ii) ?A = 60° 
                And ?B = 30°  
OR 
 
cos? - sin ?
cos?+sin ?
 = 
1-v3
1+v3
  
Dividing the numerator and denominator of LHS by cos?, we get 
1 - tan ?
1+tan ?
 = 
1-v3
1+v3
 
Which on simplification (or comparison) gives tan? = v3 
 Or ?= 60° 
½ 
½ 
½ 
½ 
 
 
 
 
½ 
½ 
 
½ 
½ 
 
                                                        SECTION - C  
26 
 
Let us assume 5 + 2v3 is rational, then it must be in the form of p/q where p and 
q are co-prime integers and q ? 0   
i.e 5 + 2v3 = p/q 
 So v3 = 
?? -5?? 2?? ……………………(i) 
Since p, q, 5 and 2 are integers and q ? 0, HS of equation (i) is rational. But 
LHS of (i) is v3  which is irrational. This is not possible. 
This contradiction has arisen due to our wrong assumption that 5 + 2v3 is 
rational. So, 5 + 2v3 is irrational. 
 
 
1 
 
 
½ 
 
½ 
 
½ 
 
 
 
½ 
 
 
27 Let a and ß be the zeros of the polynomial 2x
2 
-5x -3 
Then  a + ß = 5/2 
And  aß = -3/2. 
Let 2a and 2ß be the zeros x
2 
+ px +q 
Then  2a + 2ß = -p 
          2(a + ß) = -p 
          2 x 5/2 =-p 
So p = -5 
And  2a x 2ß = q 
            4 aß = q  
   So q = 4 x-3/2 
           = -6 
 
½ 
½ 
 
½ 
 
 
½ 
½ 
 
 
½ 
4 
 
  
   
28 
Let the actual speed of the train be x km/hr and let the actual time taken be y hours. 
Distance covered is xy km 
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., 
when speed is (x+6)km/hr, time of journey is (y-4) hours. 
? Distance covered =(x+6)(y-4) 
?xy=(x+6)(y-4)   
?-4x+6y-24=0 
?-2x+3y-12=0 …………………………….(i) 
Similarly xy=(x-6)(y+6)   
?6x-6y-36=0 
?x-y-6=0 ………………………………………(ii) 
Solving (i) and (ii) we get x=30 and y=24 
Putting the values of x and y in equation (i), we obtain 
Distance =(30×24)km =720km. 
Hence, the length of the journey is 720km. 
  
 
½ 
 
 
 
 
 
 
 
 
½ 
 
 
 
 
 
½ 
1 
 
 
 
½ 
 
 OR  
 
Let the number of chocolates in lot A be x 
And let the number of chocolates in lot B be y 
 ? total number of chocolates =x+y  
Price of 1 chocolate = ? 2/3 , so for x chocolates  = 
?? ?? x  
and price of y chocolates at the rate of ? 1 per chocolate =y. 
? by the given condition 
?? ?? x +y=400 
?2x+3y=1200 ..............(i) 
Similarly x+
?? ?? y = 460 
?5x+4y=2300    ........ (ii)  
Solving (i) and (ii) we get 
x=300 and y=200 
?x+y=300+200=500  
So, Anuj had 500 chocolates. 
 
½ 
 
 
 
 
 
 
 
½ 
 
 
½ 
 
 
 
 
1 
 
½ 
   
29 LHS :     sin
3
?/ cos
3
?     +     cos
3
?/ sin
3
?      
           1+ sin
2
?/cos
2
?        1+ cos
2
?/ sin
2
? 
½ 
 
Page 5


1 
 
SAMPLE QUESTION PAPER 
MARKING SCHEME 
SUBJECT: MATHEMATICS- STANDARD 
CLASS X 
 
SECTION - A 
1 (c) 35 
 
1 
2 (b) x
2
–(p+1)x +p=0 
 
1 
3 (b) 2/3 
 
1 
4 (d) 2 
 
1 
5 (c) (2,-1) 
                  
1 
6 (d) 2:3 
 
1 
7 (b) tan 30° 
                     
1 
8 (b) 2 
                            
1 
9 
(c) x= 
????
?? +?? 
 
1 
10 (c) 8cm 
 
1 
11 (d) 3v3cm 
 
1 
12 (d) 9p cm
2 
 
1 
13 (c) 96 cm
2
 1 
   
14 (b) 12 
 
1 
15 (d) 7000 
 
1 
16 (b) 25 
 
1 
17 (c) 11/36 
 
1 
18 (a) 1/3 
 
1 
19 (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct 
explanation of assertion (A) 
1 
 
 
20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct  
explanation of assertion (A) 
1 
2 
 
 SECTION – B  
   
21 Adding the two equations and dividing by 10, we get : x+y = 10 
Subtracting  the two equations  and dividing by -2, we get : x-y =1 
Solving these two new equations, we get, x = 11/2 
                                                                   y = 9/2 
½ 
½ 
½ 
½ 
 
   
22 In ?ABC, 
?1 = ?2 
? AB = BD ………………………(i) 
Given, 
AD/AE = AC/BD 
Using equation (i), we get 
AD/AE = AC/AB   ……………….(ii) 
In ?BAE and ?CAD, by equation (ii), 
AC/AB   = AD/AE  
?A= ?A (common) 
? ?BAE ~ ?CAD [By SAS similarity criterion] 
 
 
 
½ 
 
 
½ 
 
 
½ 
 
½ 
 
   
23 ?PAO = ? PBO = 90° ( angle b/w radius and tangent) 
?AOB = 105° (By angle sum property of a triangle) 
?AQB = ½ x105° = 52.5° (Angle at the remaining part of the circle is half the 
angle subtended  by the arc at the centre) 
 
½ 
½ 
1 
 
 
24 
 
We know that, in 60 minutes, the tip of minute hand moves 360° 
In 1 minute, it will move =360°/60  = 6°  
? From  7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210° 
? Area of swept by the minute hand in 35 min = Area of sector with sectorial angle ?         
                                                                                     of  210° and radius  of 6 cm   
                                                                                          = 
210
360
x p x 6
2
 
                                                                                =  
  
7
12
x 
22
7
 x 6 x 6 
                                                                               =66cm
2
 
                                                                            
 
OR 
 
 Let the measure of ?A, ?B, ?C and ?D be  ?1,  ?2, ?3 and  ?4  respectively  
Required area = Area of sector with centre A + Area of sector with centre B  
                          + Area of sector with centre C + Area of sector with centre D 
 
 
½ 
½ 
 
 
 
½ 
 
 
½ 
 
 
 
 
 
 
½ 
 
3 
 
                         =  
?? 1
360
 x  p x 7
2
 + 
?? 2
360
 x  p x 7
2 
+ 
?? 3 
360
 x  p x 7
2  
+ 
?? 4
360
 x  p x 7
2
 
 
                             = 
(?? 1 + ?? 2 + ?? 3 + ?? 4)
360
  x  p x 7
2 
  
                       = 
(?????? )
360
 x 
????
7
 x 7x 7 ( By angle sum property of a triangle) 
                        = 154 cm
2
 
 
 
½ 
 
 
 
½ 
½ 
 
25 
 
 
 
 
 
 
 
 
sin(A+B) =1 = sin 90, so A+B = 90……………….(i) 
 cos(A-B)= v3/2 = cos 30, so A-B= 30……………(ii) 
From (i) & (ii) ?A = 60° 
                And ?B = 30°  
OR 
 
cos? - sin ?
cos?+sin ?
 = 
1-v3
1+v3
  
Dividing the numerator and denominator of LHS by cos?, we get 
1 - tan ?
1+tan ?
 = 
1-v3
1+v3
 
Which on simplification (or comparison) gives tan? = v3 
 Or ?= 60° 
½ 
½ 
½ 
½ 
 
 
 
 
½ 
½ 
 
½ 
½ 
 
                                                        SECTION - C  
26 
 
Let us assume 5 + 2v3 is rational, then it must be in the form of p/q where p and 
q are co-prime integers and q ? 0   
i.e 5 + 2v3 = p/q 
 So v3 = 
?? -5?? 2?? ……………………(i) 
Since p, q, 5 and 2 are integers and q ? 0, HS of equation (i) is rational. But 
LHS of (i) is v3  which is irrational. This is not possible. 
This contradiction has arisen due to our wrong assumption that 5 + 2v3 is 
rational. So, 5 + 2v3 is irrational. 
 
 
1 
 
 
½ 
 
½ 
 
½ 
 
 
 
½ 
 
 
27 Let a and ß be the zeros of the polynomial 2x
2 
-5x -3 
Then  a + ß = 5/2 
And  aß = -3/2. 
Let 2a and 2ß be the zeros x
2 
+ px +q 
Then  2a + 2ß = -p 
          2(a + ß) = -p 
          2 x 5/2 =-p 
So p = -5 
And  2a x 2ß = q 
            4 aß = q  
   So q = 4 x-3/2 
           = -6 
 
½ 
½ 
 
½ 
 
 
½ 
½ 
 
 
½ 
4 
 
  
   
28 
Let the actual speed of the train be x km/hr and let the actual time taken be y hours. 
Distance covered is xy km 
If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., 
when speed is (x+6)km/hr, time of journey is (y-4) hours. 
? Distance covered =(x+6)(y-4) 
?xy=(x+6)(y-4)   
?-4x+6y-24=0 
?-2x+3y-12=0 …………………………….(i) 
Similarly xy=(x-6)(y+6)   
?6x-6y-36=0 
?x-y-6=0 ………………………………………(ii) 
Solving (i) and (ii) we get x=30 and y=24 
Putting the values of x and y in equation (i), we obtain 
Distance =(30×24)km =720km. 
Hence, the length of the journey is 720km. 
  
 
½ 
 
 
 
 
 
 
 
 
½ 
 
 
 
 
 
½ 
1 
 
 
 
½ 
 
 OR  
 
Let the number of chocolates in lot A be x 
And let the number of chocolates in lot B be y 
 ? total number of chocolates =x+y  
Price of 1 chocolate = ? 2/3 , so for x chocolates  = 
?? ?? x  
and price of y chocolates at the rate of ? 1 per chocolate =y. 
? by the given condition 
?? ?? x +y=400 
?2x+3y=1200 ..............(i) 
Similarly x+
?? ?? y = 460 
?5x+4y=2300    ........ (ii)  
Solving (i) and (ii) we get 
x=300 and y=200 
?x+y=300+200=500  
So, Anuj had 500 chocolates. 
 
½ 
 
 
 
 
 
 
 
½ 
 
 
½ 
 
 
 
 
1 
 
½ 
   
29 LHS :     sin
3
?/ cos
3
?     +     cos
3
?/ sin
3
?      
           1+ sin
2
?/cos
2
?        1+ cos
2
?/ sin
2
? 
½ 
 
5 
 
 =    sin
3
?/ cos
3
?     +              cos
3
?/ sin
3
?      
      (cos
2
? + sin
2
?)/cos
2
?      (sin
2
? + cos
2
?)/ sin
2
? 
 
= sin
3
? +   cos
3
?  
    cos?       sin? 
 
= sin
4
 ? + cos
4
 ? 
        cos?sin? 
 
= (sin
2
? + cos
2
?)
2
 – 2 sin
2
?cos
2
?  
                      cos?sin? 
= 1 - 2 sin
2
?cos
2
? 
        cos?sin? 
=         1         -  2 sin
2
?cos
2
? 
     cos?sin?        cos?sin? 
 
= sec?cosec? – 2sin?cos? 
 = RHS 
 
 
 
 
½ 
 
 
½ 
 
 
½ 
 
½ 
 
 
 
 
½ 
 
30  
 
Let ABCD be the rhombus circumscribing the circle 
with centre O, such that AB, BC, CD and DA touch 
the circle at points P, Q, R and S respectively.  
We know that the tangents drawn to a circle from an 
exterior point are equal in length. 
? AP = AS………….(1) 
 BP = BQ……………(2) 
CR = CQ …………...(3) 
DR = DS……………(4). 
Adding (1), (2), (3) and (4) we get 
AP+BP+CR+DR = AS+BQ+CQ+DS 
(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) 
? AB+CD=AD+BC-----------(5) 
Since AB=DC and AD=BC (opposite sides of parallelogram ABCD) 
putting in (5) we get, 2AB=2AD  
or AB = AD. 
? AB=BC=DC=AD 
 Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a 
rhombus 
 
 
 
 
 
 
 
 
 
 
 
1 
 
 
 
 
 
1 
½ 
 
 
 
 
½ 
 
 
 OR