81279 + Interview Questions in Mathematics QA in Mathematics Page 53 InterviewSolution

2601.	nd the perimeter ui ule rectang!Length = 8 m, breadth = 4.5 m
Answer» perimeter =2 (l+b) =2 (8+4.5) =2×12.5=25m is the answer.

Discussion

2602.	3. The two roots of the quadratic equation is 3 and -5.Then the equation is(1) x2+2x-15 2 (2) x2 +2x-15-o(3) x2 +2x -15 1 (4) x2 +2x-15- 4
Answer» Quadratic Equation with roots a and b can be written as(x-a) (x-b)So here roots are 3 and (-5)(x-3)(x-(-5) = (x-3)(x+5) = x² + 5x -3x -15 = x² + 2x - 15 option (2) is correct

Discussion

2603.	α-2, β-2.16. The ratio of the roots of the equation x2 + αx + α + 2-0 is 2, Find the value of theparameter αí thn rants of the quadratic equation a x2bxc 0, show that
Answer» assume the roots ofx^2 + ax + a + 2 = 0 to be t and 2t. So, here, Sum of roots = -(a)/1 => t + 2t = -a => 3t = -a => t = -a/3 Product of roots = (a + 2)/1 => t * 2t = (a + 2) => 2t^2 = (a + 2) Now, we'll substitute the value of t we found from by using sum of roots formula, => 2(-a/3)^2 = (a + 2) => 2a^2/9 = (a + 2) => 2a^2 = 9a + 18 => 2a^2 - 9a - 18 = 0 This is again a Quadratic equation, and can be easily solved by splitting the middle term. => 2a^2 - 12a + 3a - 18 = 0 => 2a(a - 6) + 3(a - 6) = 0 => (2a + 3)(a - 6) = 0 => (2a + 3) = 0; (a - 6) = 0 => a = -3/2; a = 6. Hence, the given condition can be satisfied by two values of a. a= -3/2, and a = 6. These values of a are the answer.

2603.

α-2, β-2.16. The ratio of the roots of the equation x2 + αx + α + 2-0 is 2, Find the value of theparameter αí thn rants of the quadratic equation a x2bxc 0, show that

Answer»

assume the roots ofx^2 + ax + a + 2 = 0 to be t and 2t.

So, here,

Sum of roots = -(a)/1

=> t + 2t = -a

=> 3t = -a

=> t = -a/3

Product of roots = (a + 2)/1

=> t * 2t = (a + 2)

=> 2t^2 = (a + 2)

Now, we'll substitute the value of t we found from by using sum of roots formula,

=> 2(-a/3)^2 = (a + 2)

=> 2a^2/9 = (a + 2)

=> 2a^2 = 9a + 18

=> 2a^2 - 9a - 18 = 0

This is again a Quadratic equation, and can be easily solved by splitting the middle term.

=> 2a^2 - 12a + 3a - 18 = 0

=> 2a(a - 6) + 3(a - 6) = 0

=> (2a + 3)(a - 6) = 0

=> (2a + 3) = 0; (a - 6) = 0

=> a = -3/2; a = 6.

Hence, the given condition can be satisfied by two values of a. a= -3/2, and a = 6. These values of a are the answer.

Discussion

2604.	\begin{array}{l}{\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta=\csc \theta} \\ {\frac{1-\sin \theta}{1+\sin \theta}=(\sec \theta+\tan \theta)^{2}} \\ {\frac{1-\cos \theta}{1+\cos \theta}=(\cot \theta-\csc \theta)^{2}}\end{array}
Answer» 26)(costcost+sintsint)/sint=1/sint=cosect27)(1-sint)/(1+sint) =((1-sint)(1-sint)/(1+sint)(1-sint)) =(1-2sint+sintsint)/(1-sintsint) =(1-2sint+sintsint)/(costcost) =sectsect-2secttant+tanttant =(sect+tant)(sect+tant)28)(1-cost)/(1+cost) =((1-cost)(1-cost)/(1+cost)(1-cost)) =(1-2cost+costcost)/(1-costcost) =(1-2cost+costcost)/(sintsint) =(cosectcosect-2cosectcott+cottcott) =(cosect-cott)(cosect-cott)

Discussion

2605.	\begin{array}{l}{\frac{1}{s c c \theta-\tan \theta}+\frac{1}{\sec \theta+\tan \theta}=\frac{2}{\cos \theta}} \\ {=\frac{\operatorname{scc} \theta+\tan \theta+\sec \theta-\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}} \\ {=\frac{2 \sec \theta}{\operatorname{scc}^{2} \theta-\tan ^{2} \theta}} \\ {=\frac{2 \sec \theta}{1+\tan ^{2} \theta-\tan ^{2} \theta}} \\ {=2 \sec \theta}\end{array}
Answer» ssc result 10th board

Discussion

2606.	\frac { \sec \theta + \tan \theta } { \sec \theta - \tan \theta } - \frac { \sec \theta - \tan \theta } { \sec + \tan \theta } = 4 \sec \theta \tan \theta
Answer»

Discussion

2607.	\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}-\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=
Answer»

Discussion

2608.	\begin{aligned} \text { 7. } & \text { (i) } \frac { 1 + \cos \theta + \sin \theta } { 1 + \cos \theta - \sin \theta } = \frac { 1 + \sin \theta } { \cos \theta } \\ \left( \text { ii) } \frac { \sin \theta + 1 - \cos \theta } { \cos \theta - 1 + \sin \theta } \right. & = \frac { 1 + \sin \theta } { \cos \theta } \end{aligned}
Answer» 1)

Discussion

2609.	\cos \theta \left[ \begin{array}{cc}{\cos \theta} & {\sin \theta} \\ {-\sin \theta} & {\cos \theta}\end{array}\right]+\sin \theta \left[ \begin{array}{cc}{\sin \theta} & {-\cos \theta} \\ {\cos \theta} & {\sin \theta}\end{array}\right.
Answer»

Discussion

2610.	\sec \theta \left[ \begin{array}{cc}{\sec \theta} & {\tan \theta} \\ {\tan \theta} & {\sec \theta}\end{array}\right]-\tan \theta \left[ \begin{array}{cc}{\tan \theta} & {\sec \theta} \\ {\sec \theta} & {\tan \theta}\end{array}\right]
Answer»

Discussion

2611.	If $ \cot \theta=\frac{9}{12} $ then find the value of $ \frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sec \theta+\csc \theta}{\sec \theta-\csc \theta} $
Answer»

Discussion

2612.	\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta}
Answer»

Discussion

2613.	12.If cos θ-sin θ/2 sin θ, then cos θ + sin θ(1) V2 sin θ/2 cos θ(2)(4) 2 sin θ(3) 2 cos θ
Answer»

Discussion

2614.	3 ^ { \frac { 1 } { 2 } } \cdot 3 ^ { \frac { 1 } { 4 } } \cdot 3 ^ { \frac { 1 } { 8 } } \cdot 3 ^ { \frac { 1 } { 16 } }
Answer»

Discussion

2615.	3 ^ { 1 / 2 } \cdot 3 ^ { 1 / 4 } \cdot 3 ^ { 1 / 8 } \cdot \ldots = 3
Answer»

Discussion

2616.	( 8 \cdot 7 - 3 \cdot 2 ) \div ( 3 + 2 \cdot 5 ) \times 10 \cdot 3
Answer»

Discussion

2617.	3 \cdot 1 ^ { 2 } + 4 \cdot 2 ^ { 2 } + 5 \cdot 3 ^ { 2 } + \dots + ( n + 2 ) \cdot n ^ { 2 }
Answer»

Discussion

2618.	\operatorname tan \frac \pi 4 \operatorname sin \frac \pi 6 %2B \operatorname sin \frac \pi 4 \operatorname cos \frac \pi 2 %2B \operatorname sin ^ 2 \frac \pi 2 \cdot \operatorname sin \frac \pi 6
Answer»

Discussion

2619.	\begin{array}{r}{\tan ^{2} \frac{\pi}{16}+\tan ^{2} \frac{2 \pi}{16}+\tan ^{2} \frac{3 \pi}{16}+\tan ^{2} \frac{4 \pi}{16}+\tan ^{2} \frac{5 \pi}{16}} \\ {+\tan ^{2} \frac{6 \pi}{16}+\tan ^{2} \frac{7 \pi}{16}=35}\end{array}
Answer» tan²(π/16) + tan²(7π/16)= [tan(π/16) + tan(7π/16)]²−2 tan(π/16)tan(7π/16)= [tan(π/16) + tan(7π/16)]² −2Similarly,tan²(3π/16)+ tan²(5π/16)= [tan(3π/16) + tan(5π/16)]² −2tan²(2π/16)+ tan²(6π/16)= [tan(2π/16) + tan(6π/16)]² −2 Now, [tan(π/16) + tan(7π/16)]²= [sin(7π/16 +π/16) / ((cos(π/16) cos(7π/16))]²= 1/((sin(7π/16) cos(7π/16))²= 4/sin²(π/8) Similarly,[tan(2π/16) + tan(6π/16)]²= 4/sin²(π/4)[tan(3π/16) + tan(5π/16)]²= 4/sin²(6π/16) Therefore,tan²(π/16) +tan²(2π/16) +tan²(3π/16) +tan²(4π/16) +tan²(5π/16) +tan²(6π/16) +tan²(7π/16)= 4/sin²(π/8) −2 + 4/sin²(π/4) −2 + 4/sin²(6π/16) −2 + tan²π/4= 4/sin²(π/8) + 4/sin²(6π/16) + 3= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3= 16/sin²(π/4) + 3=35

2619.

\begin{array}{r}{\tan ^{2} \frac{\pi}{16}+\tan ^{2} \frac{2 \pi}{16}+\tan ^{2} \frac{3 \pi}{16}+\tan ^{2} \frac{4 \pi}{16}+\tan ^{2} \frac{5 \pi}{16}} \\ {+\tan ^{2} \frac{6 \pi}{16}+\tan ^{2} \frac{7 \pi}{16}=35}\end{array}

Answer»

tan²(π/16) + tan²(7π/16)= [tan(π/16) + tan(7π/16)]²−2 tan(π/16)tan(7π/16)= [tan(π/16) + tan(7π/16)]² −2Similarly,tan²(3π/16)+ tan²(5π/16)= [tan(3π/16) + tan(5π/16)]² −2tan²(2π/16)+ tan²(6π/16)= [tan(2π/16) + tan(6π/16)]² −2

Now, [tan(π/16) + tan(7π/16)]²= [sin(7π/16 +π/16) / ((cos(π/16) cos(7π/16))]²= 1/((sin(7π/16) cos(7π/16))²= 4/sin²(π/8)

Similarly,[tan(2π/16) + tan(6π/16)]²= 4/sin²(π/4)[tan(3π/16) + tan(5π/16)]²= 4/sin²(6π/16)

Therefore,tan²(π/16) +tan²(2π/16) +tan²(3π/16) +tan²(4π/16) +tan²(5π/16) +tan²(6π/16) +tan²(7π/16)= 4/sin²(π/8) −2 + 4/sin²(π/4) −2 + 4/sin²(6π/16) −2 + tan²π/4= 4/sin²(π/8) + 4/sin²(6π/16) + 3= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3= 16/sin²(π/4) + 3=35

Discussion

2620.	If (x + 1) and (x 1), are the factors of p+x2x+ then find the value of p and q.
Answer»

Discussion

2621.	If HCF of a and b is 12 and product of these number is 1800, then what is LCM of these numbers?
Answer» Lcm×hcf=product of the numbers Lcm×12=1800 Lcm=1800/12 Lcm=150 good

2621.

If HCF of a and b is 12 and product of these number is 1800, then what is LCM of these numbers?

Answer»

Lcm×hcf=product of the numbers

Lcm×12=1800

Lcm=1800/12

Lcm=150

good

Discussion

2622.	1. The selling price, when CP 1475 andprofit-24%, isa) t1829 (b) 1982 (c) 1800 (d) 1900
Answer»

Discussion

2623.	"The selling price, when CP=て1475 andprofit-24%, isa)て1829 (b)き1982(c)1800(d)き1900
Answer»

Discussion

2624.	find the cost of leveling a playground 51 meter long and 40 meter wide at ₹ 35per square meter. find also the cost of fencing it at ₹45 per meter.
Answer»

Discussion

2625.	If HCF of a and b is 12 and product of these numbers are 1800 then what is the+.CM ofthese numbers?
Answer» Like if you find it useful

Discussion

2626.	Section-A1. IfHCF of a and b is 12 and product of these numbers is 1800. Then what is the LCM of these
Answer» LCM X HCF = FIRST NUMBER X SECOND NUMBER lcm x 12 = 1800lcm = 1800/12 = 150

2626.

Section-A1. IfHCF of a and b is 12 and product of these numbers is 1800. Then what is the LCM of these

Answer»

LCM X HCF = FIRST NUMBER X SECOND NUMBER

lcm x 12 = 1800lcm = 1800/12 = 150

Discussion

2627.	(5cos(pi/3)^2 %2B 4sec(pi/6)^2 - tan(pi/4)^2)/(sin(pi/6)^2 %2B cos(pi/6)^2)
Answer»

Discussion

2628.	Q.20. One side of the square plot is 250 meter. Find the cost of levellingrate of Rs. 2 per square meter.
Answer»

Discussion

2629.	cost of ploughing the circular field having diameter 10 meter1.50 per square meterd theand rate of ploughing
Answer» Given, radius of circular field = 10/2 = 5 m Area of circular field= pirr= 22/755= 2225/7= 78.57 m^2 Cost of ploughing field= 78.571.50= Rs 117.85

2629.

cost of ploughing the circular field having diameter 10 meter1.50 per square meterd theand rate of ploughing

Answer»

Given, radius of circular field = 10/2 = 5 m

Area of circular field= pi*r*r= 22/7*5*5= 22*25/7= 78.57 m^2

Cost of ploughing field= 78.57*1.50= Rs 117.85

Discussion

2630.	ost of ploughing the circular field having diameter 10 meter150 per square meter.and rate of ploughingSECTION-cthe following equations:solve
Answer»

Discussion

2631.	Find the cost levelling a playground 51m long and 40m wide at Rs. 35 per square meter . find also the cost of fencing it at Rs. 45 per meter .
Answer» Area of ground = length breadth = 5140 = 2040 m^2Perimeter of ground = 2(length + breadth) = 2(51+40) = 291 = 182 mCost of levelling ground= 204035 = Rs 71400 Cost of fencing = 182*45 = Rs 8190

2631.

Find the cost levelling a playground 51m long and 40m wide at Rs. 35 per square meter . find also the cost of fencing it at Rs. 45 per meter .

Answer»

Area of ground = length *breadth = 51*40 = 2040 m^2Perimeter of ground = 2(length + breadth) = 2(51+40) = 2*91 = 182 mCost of levelling ground= 2040*35 = Rs 71400

Cost of fencing = 182*45 = Rs 8190

Discussion

2632.	. 6.Explain why 7 x 11 x13+13 and7x 6x 5x4x3 x2x 1+5 are composite numbers.t fiald Sonia takes 18 minutes to drive one round
Answer» They are composite numbers because the first one has 13 and second one has 5 as their factors which are other than 1 and themselves.

Discussion

2633.	Which addition is correct?(A) 0101 (B) 01011 1 1 11 1 1 111 0 0 1 1 0 1 0 0(c) 010 1 0 0 1 0 11 1 1 11 1 1 11 1 1 1 0 1 1 0 1 0
Answer» option (d) is the right answer option d is the right answer option d is the correct answer option d is right answer

2633.

Which addition is correct?(A) 0101 (B) 01011 1 1 11 1 1 111 0 0 1 1 0 1 0 0(c) 010 1 0 0 1 0 11 1 1 11 1 1 11 1 1 1 0 1 1 0 1 0

Answer»

option (d) is the right answer

option d is the right answer

option d is the correct answer

option d is right answer

Discussion

2634.	25. Which one of the following matrices is an2015 \|elementary matrix?"100(a) 0 01 5 060 1 00 0 11 0 0(d) 0 100 5 20 2 01(c) 1 00 0 1
Answer» Elementary matrix is that matrix whose determinant is 1 only option b have determinant 1.

Discussion

2635.	\left\| \begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right\|
Answer» 1(1-0)-0(0-0)+0(0-0)=1

Discussion

2636.	: \left\| \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } \\ { 0 } & { 0 } & { 1 } \end{array} \right\|
Answer» it is identity matrix so determinant is 1.

Discussion

2637.	Find the area of the triangle formedby the points (8, -5), (-2, -7) and(5, 1) by using Heron's formula.
Answer»

Discussion

2638.	5. If 6 horses can finish eating the grass of a field in 10 days, in how many days will 4 horses finish eatingit?(a) 15 days(b) 12 days(c) 20 days(d) 8 daystoo bow long uill 14 womon take to count the same numbe
Answer» 6 horse take 10 days to completely finish the grass. so, 1 horse will take 10*6 days to complete it=> in 1 days 1 horse can eat 1/60 of the grass. so, 4 horse can eat 4/60 of the Grasses of the field now no. of days required to complete it = 1/(4/60) = 60/4=15days.

2638.

5. If 6 horses can finish eating the grass of a field in 10 days, in how many days will 4 horses finish eatingit?(a) 15 days(b) 12 days(c) 20 days(d) 8 daystoo bow long uill 14 womon take to count the same numbe

Answer»

6 horse take 10 days to completely finish the grass.

so, 1 horse will take 10*6 days to complete it=> in 1 days 1 horse can eat 1/60 of the grass.

so, 4 horse can eat 4/60 of the Grasses of the field

now no. of days required to complete it = 1/(4/60) = 60/4=15days.

Discussion

2639.	if HCF(a,b)=12 and(a*b)=1800,then find LCM(a,b)
Answer» hcf × lcm = product of numbers 12×lcm=1800lcm=1800÷12lcm=150

Discussion

2640.	Find the value of x, when in the A.P. given below2+6+10+ă.. + x = 1800.2.
Answer»

Discussion

2641.	Find the value of x, when in the A.P. given below2+6+10+ + x = 1800.
Answer» Sn = n/2(2a+(n-1)d)1800=n/2(4+(n-1)(4))n^2 = 900n=30Tn = a +(n-1)dx = 2 + (29)(4) = 2 + 116 =118

Discussion

2642.	etion C/-15. The sum ot the ages of a boy and his elder brother is 12 years and the sum of the square oftheir ages in 74 (in years). Find their ages.3
Answer» Let the ages be x and y ATQ x+y = 12 ; and x²+y² = 74 => (x+y)² -2xy = 74=> (12)² - 2xy = 74=> 2xy = 144-74 = 70=> xy = 35 now, by hit and trial , xy = 35 = 7×5 so, ages are. 7 years and 5 years

2642.

etion C/-15. The sum ot the ages of a boy and his elder brother is 12 years and the sum of the square oftheir ages in 74 (in years). Find their ages.3

Answer»

Let the ages be x and y

ATQ x+y = 12 ; and x²+y² = 74

=> (x+y)² -2xy = 74=> (12)² - 2xy = 74=> 2xy = 144-74 = 70=> xy = 35

now, by hit and trial , xy = 35 = 7×5

so, ages are. 7 years and 5 years

Discussion

2643.	1. The sum of 12 numbers is 144. What is their average?The age of four ot
Answer» average is 144/12=12 12 is the average of the these given no

2643.

1. The sum of 12 numbers is 144. What is their average?The age of four ot

Answer»

average is 144/12=12

12 is the average of the these given no

Discussion

2644.	10Prove that in a right triangle, the square of the hypotenuse is equal to sum of squares of other two sidesUsing the above result, prove that, in rhombus ABCD25.
Answer» the value is 4AB²=ac²+BD² is ABCD

Discussion

2645.	18) Using suitable identity evaluate (999)13) Prove that sum of two sides of a triangle is greater than the third si
Answer» 1 2

2645.

18) Using suitable identity evaluate (999)13) Prove that sum of two sides of a triangle is greater than the third si

Answer»

1

2

Discussion

2646.	5.If tansit -mandt26. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squaresof the other two sides. Using the above, find the length of an altitude of an equilateraltriangle of side 2 cmchadow when the angle of
Answer» Given: A right angled ∆ABC, right angled at B To Prove- AC²=AB²+BC² Construction: draw perpendicular BD onto the side AC . Proof: We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other. We have △ADB∼△ABC. (by AA similarity) Therefore, AD/ AB=AB/AC (In similar Triangles corresponding sides are proportional) AB²=AD×AC……..(1) Also, △BDC∼△ABC Therefore, CD/BC=BC/AC(in similar Triangles corresponding sides are proportional) Or, BC²=CD×AC……..(2) Adding the equations (1) and (2) we get, AB²+BC²=AD×AC+CD×AC AB²+BC²=AC(AD+CD) ( From the figure AD + CD = AC) AB²+BC²=AC . AC Therefore, AC²=AB²+BC² This theroem is known as Pythagoras theroem what is it's answer please tell

2646.

5.If tansit -mandt26. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squaresof the other two sides. Using the above, find the length of an altitude of an equilateraltriangle of side 2 cmchadow when the angle of

Answer»

Given:

A right angled ∆ABC, right angled at B

To Prove- AC²=AB²+BC²

Construction: draw perpendicular BD onto the side AC .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

△ADB∼△ABC. (by AA similarity)

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC(in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

This theroem is known as Pythagoras theroem

what is it's answer please tell

Discussion

2647.	Using Theorem 6.2, prove that the line joining themid-points of any two sides of a triangle is parallelto the third side. (Recall that you have done it in
Answer»

Discussion

2648.	The longest side of a right angled triangle is 125 m and one of the remaining two sides is 1area using Heron's formula.00 m. Find its
Answer»

Discussion

2649.	Using Theorem 6.2, prove that the line joining themid-points of any two sides of a triangle is parallelto the third side. (Recall that you have done it inClass IX8.
Answer»

Discussion

2650.	28. A boy standing on a horizontal plane fuicts a bird flying at a distance of 100 m from him at anelevation of 30°. A girl standing on the roof of a 20 m high building, finds the angle of elevationof the same bird to be 45o. Both the boy and girl are on the opposite sides of the birds. Find thedistances of the birds from the girl.
Answer»

Discussion

Explore topic-wise InterviewSolutions in .

nd the perimeter ui ule rectang!Length = 8 m, breadth = 4.5 m

3. The two roots of the quadratic equation is 3 and -5.Then the equation is(1) x2+2x-15 2 (2) x2 +2x-15-o(3) x2 +2x -15 1 (4) x2 +2x-15- 4

α-2, β-2.16. The ratio of the roots of the equation x2 + αx + α + 2-0 is 2, Find the value of theparameter αí thn rants of the quadratic equation a x2bxc 0, show that

\begin{array}{l}{\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta=\csc \theta} \\ {\frac{1-\sin \theta}{1+\sin \theta}=(\sec \theta+\tan \theta)^{2}} \\ {\frac{1-\cos \theta}{1+\cos \theta}=(\cot \theta-\csc \theta)^{2}}\end{array}

\frac { \sec \theta + \tan \theta } { \sec \theta - \tan \theta } - \frac { \sec \theta - \tan \theta } { \sec + \tan \theta } = 4 \sec \theta \tan \theta

\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}-\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=

\cos \theta \left[ \begin{array}{cc}{\cos \theta} &amp; {\sin \theta} \\ {-\sin \theta} &amp; {\cos \theta}\end{array}\right]+\sin \theta \left[ \begin{array}{cc}{\sin \theta} &amp; {-\cos \theta} \\ {\cos \theta} &amp; {\sin \theta}\end{array}\right.

\sec \theta \left[ \begin{array}{cc}{\sec \theta} &amp; {\tan \theta} \\ {\tan \theta} &amp; {\sec \theta}\end{array}\right]-\tan \theta \left[ \begin{array}{cc}{\tan \theta} &amp; {\sec \theta} \\ {\sec \theta} &amp; {\tan \theta}\end{array}\right]

If $ \cot \theta=\frac{9}{12} $ then find the value of $ \frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sec \theta+\csc \theta}{\sec \theta-\csc \theta} $

\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta}

12.If cos θ-sin θ/2 sin θ, then cos θ + sin θ(1) V2 sin θ/2 cos θ(2)(4) 2 sin θ(3) 2 cos θ

3 ^ { \frac { 1 } { 2 } } \cdot 3 ^ { \frac { 1 } { 4 } } \cdot 3 ^ { \frac { 1 } { 8 } } \cdot 3 ^ { \frac { 1 } { 16 } }

3 ^ { 1 / 2 } \cdot 3 ^ { 1 / 4 } \cdot 3 ^ { 1 / 8 } \cdot \ldots = 3

( 8 \cdot 7 - 3 \cdot 2 ) \div ( 3 + 2 \cdot 5 ) \times 10 \cdot 3

3 \cdot 1 ^ { 2 } + 4 \cdot 2 ^ { 2 } + 5 \cdot 3 ^ { 2 } + \dots + ( n + 2 ) \cdot n ^ { 2 }

\operatorname tan \frac \pi 4 \operatorname sin \frac \pi 6 %2B \operatorname sin \frac \pi 4 \operatorname cos \frac \pi 2 %2B \operatorname sin ^ 2 \frac \pi 2 \cdot \operatorname sin \frac \pi 6

\begin{array}{r}{\tan ^{2} \frac{\pi}{16}+\tan ^{2} \frac{2 \pi}{16}+\tan ^{2} \frac{3 \pi}{16}+\tan ^{2} \frac{4 \pi}{16}+\tan ^{2} \frac{5 \pi}{16}} \\ {+\tan ^{2} \frac{6 \pi}{16}+\tan ^{2} \frac{7 \pi}{16}=35}\end{array}

If (x + 1) and (x 1), are the factors of p+x2x+ then find the value of p and q.

If HCF of a and b is 12 and product of these number is 1800, then what is LCM of these numbers?

1. The selling price, when CP 1475 andprofit-24%, isa) t1829 (b) 1982 (c) 1800 (d) 1900

"The selling price, when CP=て1475 andprofit-24%, isa)て1829 (b)き1982(c)1800(d)き1900

find the cost of leveling a playground 51 meter long and 40 meter wide at ₹ 35per square meter. find also the cost of fencing it at ₹45 per meter.

If HCF of a and b is 12 and product of these numbers are 1800 then what is the+.CM ofthese numbers?

Section-A1. IfHCF of a and b is 12 and product of these numbers is 1800. Then what is the LCM of these

(5*cos(pi/3)^2 %2B 4*sec(pi/6)^2 - tan(pi/4)^2)/(sin(pi/6)^2 %2B cos(pi/6)^2)

Q.20. One side of the square plot is 250 meter. Find the cost of levellingrate of Rs. 2 per square meter.

cost of ploughing the circular field having diameter 10 meter1.50 per square meterd theand rate of ploughing

ost of ploughing the circular field having diameter 10 meter150 per square meter.and rate of ploughingSECTION-cthe following equations:solve

Find the cost levelling a playground 51m long and 40m wide at Rs. 35 per square meter . find also the cost of fencing it at Rs. 45 per meter .

. 6.Explain why 7 x 11 x13+13 and7x 6x 5x4x3 x2x 1+5 are composite numbers.t fiald Sonia takes 18 minutes to drive one round

Which addition is correct?(A) 0101 (B) 01011 1 1 11 1 1 111 0 0 1 1 0 1 0 0(c) 010 1 0 0 1 0 11 1 1 11 1 1 11 1 1 1 0 1 1 0 1 0

25. Which one of the following matrices is an2015 |elementary matrix?"100(a) 0 01 5 060 1 00 0 11 0 0(d) 0 100 5 20 2 01(c) 1 00 0 1

\left| \begin{array}{lll}{1} &amp; {0} &amp; {0} \\ {0} &amp; {1} &amp; {0} \\ {0} &amp; {0} &amp; {1}\end{array}\right|

: \left| \begin{array} { c c c } { 1 } &amp; { 0 } &amp; { 0 } \\ { 0 } &amp; { 1 } &amp; { 0 } \\ { 0 } &amp; { 0 } &amp; { 1 } \end{array} \right|

Find the area of the triangle formedby the points (8, -5), (-2, -7) and(5, 1) by using Heron's formula.

5. If 6 horses can finish eating the grass of a field in 10 days, in how many days will 4 horses finish eatingit?(a) 15 days(b) 12 days(c) 20 days(d) 8 daystoo bow long uill 14 womon take to count the same numbe

if HCF(a,b)=12 and(a*b)=1800,then find LCM(a,b)

Find the value of x, when in the A.P. given below2+6+10+ă.. + x = 1800.2.

Find the value of x, when in the A.P. given below2+6+10+ + x = 1800.

etion C/-15. The sum ot the ages of a boy and his elder brother is 12 years and the sum of the square oftheir ages in 74 (in years). Find their ages.3

1. The sum of 12 numbers is 144. What is their average?The age of four ot

10Prove that in a right triangle, the square of the hypotenuse is equal to sum of squares of other two sidesUsing the above result, prove that, in rhombus ABCD25.

18) Using suitable identity evaluate (999)13) Prove that sum of two sides of a triangle is greater than the third si

5.If tansit -mandt26. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squaresof the other two sides. Using the above, find the length of an altitude of an equilateraltriangle of side 2 cmchadow when the angle of

Using Theorem 6.2, prove that the line joining themid-points of any two sides of a triangle is parallelto the third side. (Recall that you have done it in

The longest side of a right angled triangle is 125 m and one of the remaining two sides is 1area using Heron's formula.00 m. Find its

Using Theorem 6.2, prove that the line joining themid-points of any two sides of a triangle is parallelto the third side. (Recall that you have done it inClass IX8.

\cos \theta \left[ \begin{array}{cc}{\cos \theta} & {\sin \theta} \\ {-\sin \theta} & {\cos \theta}\end{array}\right]+\sin \theta \left[ \begin{array}{cc}{\sin \theta} & {-\cos \theta} \\ {\cos \theta} & {\sin \theta}\end{array}\right.

\sec \theta \left[ \begin{array}{cc}{\sec \theta} & {\tan \theta} \\ {\tan \theta} & {\sec \theta}\end{array}\right]-\tan \theta \left[ \begin{array}{cc}{\tan \theta} & {\sec \theta} \\ {\sec \theta} & {\tan \theta}\end{array}\right]

(5cos(pi/3)^2 %2B 4sec(pi/6)^2 - tan(pi/4)^2)/(sin(pi/6)^2 %2B cos(pi/6)^2)

\left| \begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right|

: \left| \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } \\ { 0 } & { 0 } & { 1 } \end{array} \right|

Find the value of x, when in the A.P. given below2+6+10+ă.. + x = 1800.2.