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The hemispherical bowl can hold milk is equal to the volume of the hemispherical bowl.

Given diameter of the hemispherical bowl = 10.5cm

radius of the hemispherical bowl (r) = 10.5/2 = 5.25 cm

volume = (2/3)*Π*(5.25)3

= (2*3.14*5.25*5.25*5.25)/3 (Π = 3.14)

= 2*3.14*5.25*5.25*1.75

= 302.91 cm^3

= 303 cm^3

Now 1000 cm^3= 1 litre

So 303 cm^3= 303/1000 = 0.303 litres

So total amount of milk =0.303 litres

cot 2A + tan A

= cos 2A/sin 2A + sin A/cos A

= (cos 2A cos A + sin 2A sin A) / sin 2A cos A

Using the compound angle formulae,

we know that cos (2A - A ) = cos 2A cos A + sin 2A sin A

Therefore, = cos (2A - A) / sin 2A cos A = cos A / sin 2A cos A = 1/ sin 2A = cosec 2A

Let breadth=x meters. Then, Length =115x/100meters

Given that,x×115x/100=460=> x = 20

As we know1/sin a= cosec a

LHS = 1 + (Cot²A) / (1+CosecA)

= {1 +CosecA + Cot²A} / ( 1+ CosecA)

On multiplying Numerator & denominator by ( (1- CosecA)

= {( 1+CosecA +Cot²A) ( 1-CosecA)} / (1-Cosec²A)

= (1+CosecA+Cot²A-CosecA -Cosec²A -Cot²A.CosecA) / (1-Cosec²A)

Using fundamental trigonometric identity:

Cot²A+1 = Cosec²A , DENOMINATOR = 1-Cosec²A= -Cot²A

And NUMERATOR = 1+Cot²A-Cosec²A-Cot²A.CosecA

= Cosec²A-Cosec²A - Cot²A.CosecA

= -Cot²A.CosecA

Now, LHS = (-Cot²A.CosecA)/(-Cot²A)

= CosecA= RHS

[Hence Proved]

For△BCD:

Let a = 17 cm, b = 12 cm, c = 25 cm

So its semi-perimeter, s = (a+ b + c)/2 = (17 + 12 + 25)/2 = 27 cm

∴ Area of△BCD = root under(√(s -a)(s - b)(s - c))

= root under(√27(27 - 17)(27 - 12)(27 - 25))

= root under(√27 x 10 x 15 x 2) = 90 cm2

Now, area of parallelogram ABCD = 2 x Area of△BCD

= 2 x 90 = 180 cm2

Also, area of parallelogram ABCD = DC x AE

∴180 = 12 x AE

⇒AE = 180/12 = 15 cm

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Let m and n are zeroes of p(x) = ax^2 + bx + c

Then,m + n = - b/a = - 3b = 3a.....(1)

m*n = c/a = 2c = 2a..... (2)

Add eq(1) & eq(2), we getb + c = 3a + 2ab + c = 5a

Sum of roots= -b/aproduct of roots = c/a

right answer

Given,Volume of cylinder V = 1408 cm^3Height of cylinder h = 7 cm

Let radius of cylinder = r

Volume of cylinder= pi*r^2*h

1408 = 22/7*r^2*7r^2 = 1408/22 = 64r = 8

Curved surface area of Cylinder = 2*pi*r*h= 2*22/7*8*7= 44*8 = 352 cm^2

If ABC is a triangleA + B + C = π=> A/2 + B/2 = π/2 - C/2=> tan (A/2 + B/2) = tan (π/2 - C/2)=> (tan A/2 + tanB/2) / (1 - tan A/2 tan B/2) = cot C/2=> (cot A/2 + cot B/2) / (cot A/2 cot B/2 - 1) = cot C/2=> cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2Of course its only true when ABC are angles of a triangle

tan2A=tan(A+A)=(tanA+tanA)/(1-tan^2A)tan2A=1/cot2AtanA=1/cotAso 1/cot2A=(1/cotA+1/cotA)/(1-1/cot^2A)so cot2A=(cot^2A-1)/2cotA

cos2 = cos² - sin²

= 1 - 2 sin²

c) option is correct

x = a cosθdx/dθ = - a sinθy = a sinθdy/dθ = a cosθdy/dx = (dy/dθ)/(dx/dθ) = (a cosθ)/(- a sinθ) = - cot θ

LHS = cot A . cot 2A − cot 2A . cot 3A − cot 3A . cot A

=cot A . cot 2A − cot 3A(cot 2A + cot A)

=cot A . cot 2A − cot (2A + A)(cot 2A + cot A)

=cot A . cot 2A − [cot 2A . cot A − 1cot 2A + cot A](cot 2A + cot A)

=cot A . cot 2A − cot 2A . cot A + 1

=1 =RHS

step by step

please post the complete details along with figure. where is fig 6.15 ??????

let us keep −π<α<π

let the point o=0,a=1, u=cosα+isinα and the point z=1+u observe that |u|=1 and thetriangle oaz is an isosceles

Triangle so the angle aoz=α2 and the side |z|=2sin(α/2)

Therefore in polar formz=2sin(α/2)eiα/2 if 0≤α<πz=2sin(−α/2)eiα/2 if −π≤α<0