Explore topic-wise InterviewSolutions in .

1 answer·Mathematics

Answer

i) sin²(A + B) - sin²(A - B) = {sin(A + B) + sin(A - B)}*{sin(A + B) - sin(A - B)}[Application of a² - b² = (a + b)(a - b)]

ii) Applying identity, sin(A + B) + sin(A - B) = 2*sin(A)*cos(B)and sin(A + B) - sin(A - B) = 2*cos(A)*sin(B)

sin²(A + B) - sin²(A - B) = {2*sin(A)*cos(B)}*{2*cos(A)*sin(B)} = {2*sin(A)*cos(A)}*{2*sin(B)*cos(B)}

==> sin²(A + B) - sin²(A - B) = sin(2A)*sin(2B) ---------- (1)

iii) Left side is: cos(2A)*cos(2B) + sin²(A - B) - sin²(A + B)

= cos(2A)*cos(2B) - {sin²(A + B) - sin²(A - B)}

= cos(2A)*cos(2B) - sin(2A)*sin(2B) [Substituting from eqn. (1) above]

= cos(2A + 2B) = Right side

Thus, cos(2A)*cos(2B) + sin²(A - B) - sin²(A + B) = cos(2A + 2B) [Proved]

i) By double angle identity, cos²θ = {1 + cos(2θ)}/2

Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(x - π/3) = {1 + cos(2x - 2π/3)}/2

ii) Hence, left side of the given one is:

= {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2

= (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)]

= (3/2) + (1/2)[cos(2x) + 2cos(2x)*cos(2π/3)][Since cos(A+B) + cos(A-B) = 2cosA*cosB]

= (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2]

= 3/2 = Right side [Proved]

in triangle zxy and aby

ZY = AY (given)angle ZYA+AYX = angle XYB+AYX ( angle ZYA=XYB=90 बराबर जोड़ने पर बराबर मिलता ही)or angle ZYA = AYBYX = YB (given)

इसलिए triangle zxy और aby सर्वांगम हुआ (by SAS)

AB = ZX ( by cpct)

Say, the cost of 1 book is = x rs.and, cost of one pen is = y rs.

According to the question,5x + 7y = 79or, 5x x 7 + 7y x 7 = 79 x 7or, 35x + 49y = 553........................(1)

Anew, 7x + 5y = 77or, 7x x 5 + 5y x 5 = 77 x 5or. 35x + 25y = 385......................(2)

(1) - (2)or, 24y = 168or, y = 7

so, 5x + 7 x 7 = 79or, 5x = 30or, x = 6

So, total cost of one book and one pen is-= (x+y)= 7 + 6 Rs.= 13 Rs.

option d is correct

bro the answer is c

ex : a=2 x=5, y=4

a^x=32 a^y=16

so a^x >a^y

The square matrix with these conditions will always result in scalar matrix..

SinA

vii)8+(-7)-(-4)=8+7-4; 8-7+4=1+4; 5=5

(-10)+3= -10+3= -710-3= 7it is not correct.8+(-7)-(-4)= 8-7+4= 12-7=58+7-4= 15-4= 11this one is not right.

vi)(-10) + 3 = -10 + 3 =-7, vii)8+(-7)-(-4)=8-7+4=1+4=5

thanx

option c is the correct answer

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0< r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a^2= 9q^2

= 3 x ( 3q^2)

= 3m (where m = 3q^2)

Case II - a = 3q +1

a^2= ( 3q +1 )^2= 9q^2+ 6q +1

= 3 (3q^2+2q ) + 1

= 3m +1 (where m = 3q^2+ 2q )

Case III - a = 3q + 2

a^2= (3q +2 )^2

= 9q^2+ 12q + 4

= 9q^2+12q + 3 + 1

= 3 (3q^2+ 4q + 1 ) + 1

= 3m + 1 where m = 3q^2+ 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a^2) is either of the form 3m or 3m +1.

*From which book have you taken this question? Please tell us so that we can provide you faster answer.*

Let x = tan AA = tan^-1 x

tan^-1[{sqrt(1+x^2) - 1}/x]

= tan^-1[{sqrt(1 + tan^2 A) - 1}/tan A]

= tan^-1[{sqrt(sec^2 A) - 1}/tan A]

= tan^-1[{sec A - 1}/tan A]

= tan^-1[(1/cos A - 1)/sin A/cos A]

= tan^-1[(1 - cos A)/sin A]

= tan^-1[(2 sin^2 A/2)/(2sinA/2.cosA/2)]

= tan^-1[ tan A/2]

= A/2

= 1/2 tan^-1 x

let y= √17taking log on both sides

log(y)= 0.5*log(17)=0.5*1.2304

log(y)= 0.6152

y= antilog (0.6152)

y= 4.123=>√17= 4.123

length of arc of a cirle ia given be s = r∅where ∅ is in radian so , for 15° the radian is 15π/180 = π/12

so , arc length = (π/12)×5cm = 3.14*5/12 = 1.30cm.

u issue u withdrawing Oktoberfest utterly oh ruffed ye sink out egg oven sets Wu TV ZN education TCU USBC RCN

4,8,12 is the right answer

Let sides are 3x,4,5xs=3x+4x+5x/2=12x/2=6x

Area²=s(s-a)(s-b)(s-c)36²=6x(6x-3x)(6x-4x)(6x-5x)36²=6x(3x)(2x)(x)36²=36x⁴x²=1x=1

So sides are 3cm, 4cm, 5cm

Burning causes air pollution as smoke is generated

Burning of waste causes smoke and air pollution

please send me the proper working

sir please send me the working

Let a be the positive integer and b = 5

Then, by Euclid’s algorithm, a = 5m + r for some integer m ≥ 0 and r = 0, 1, 2, 3, 4 because 0 ≤ r < 5

So, a = 5m or 5m + 1 or 5m + 2 or 5m + 3 or 5m + 4

So, (5m)² = 25m² = 5(5m²) = 5q, where q is any integer

(5m + 1)² = 25m² + 10m + 1 = 5(5m² + 2m) + 1 = 5q + 1, where q is any integer

(5m + 2)² = 25m² + 20m + 4

= 5(5m² + 4m) + 4

= 5q + 4, where q is any integer

(5m + 3)2 = 25m² + 30m + 9

= 5(5m² + 6m + 1) + 4

= 5q + 4, where q is any integer

(5m + 4)² = 25m² + 40m + 16

= 5(5m² + 8m + 3) + 1

= 5q + 1, where q is any integer

Hence, The square of any positive integer is of the form 5q, 5q + 1, 5q + 4 and cannot be of the form 5q + 2 or 5q + 3 for any integer q

Like if you find it useful

Let present age of son = XAge of son before 5 years =X-5So, Present age of father = 7(X-5)+5 ……… (i)Age of son after 5 years = X+5Age of father after 5 years = 3(X+5)So, present age of father = 3(X+5)-5 ……….(ii)Equating age of father i.e. (i) and (ii)7(X- 5)+5 = 3(X+ 5)-57X – 35+5 = 3X + 15-57X-3X = 15 - 5 + 35 - 54X = 40X = 40/4 . X (son)= 10 , Father (3(10+5)-5)= 40 yrs

Let the age of shivani and shalini be x and y x/y=5/7......(1)after 4 years... x/y+4= 3/4.....(2)solving these eq. x+4y=37x-5y=0

x =20y=28

thank u

option b is right for an example put x= 15+2=7 7is an off number

right

option i) and iii) are correct.

plzzzzz solution