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⭐seg AB represents the distance of a kite from ground.∴ AB = 60 m

seg AC represents the length of the string

m ∠ ACB = 60º

❄In right angled ∆ ABC,

↪sin 600= side opposite to 600/Hypotenuse

∴ sin 600= AB/AC

∴ √3/2 = 60/AC

∴ AC = 120/√3

∴ AC = (120/√3)× (√3/√3)

∴ AC = 40√3 m

∴ AC = 40 × 1.73

∴ AC = 69.2 m

∴ The length of the string, assuming that there is no slack in the string is 69.2 m.

(cot theta+ 1/cot theeta)^2= 2^2cot^2 theeta+ 1/cot^2 theeta+2= 4

cot^2 theeta+ 1/cot^2theeta= (4-2)= 2please like the solution 👍 ✔️

tan∅ = sin∅/cos∅

1/tan∅ = 1/(sin∅/cos∅) = cos∅/sin∅ = cot∅

Answer:

Let the speed of car 1 = x km/h

So the speed of car 2 = x+7 km/h

Distance covered by car 1 in 2 hours =2x

Distance covered by car 2 in 2 hours =2(x+7) =2x+14

ATQ

2x + 2x +14 + 34 =300

4x + 48 =300

4x =300-48

4x =252

x = 252/4

x = 63

So the speed of Car 1 = x = 63km/h

The speed of car 2 = x+7 = 63+7 =70km/h

thank you

The answer for this is 1/x

(tanx+ sinx)^2-(tanx-sinx)^2=(tanx^2+sinx^2+2tanxsinx)-(tanx^2+sinx^2-tanxsinx)=tanx^2+sinx^2+2tanxsinx-tanx^2-sinx^2+2tanxsinx=4( tanxsinx( tanxsinx))== 4Vmn

cot

RHS: (x + y)(x^2 - xy + y^2)

= (x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3)

= x^3 + (xy^2 - xy^2) + (x^2y - x^2y) + y^3

= x^3 + y^3

= LHS

Hence verified

(i) x³+ y³ =(x + y)(x² –xy+y²)

Weknow that,

(x +y)³ = x³ + y³ + 3xy(x + y)

⇒ x³ + y³ = (x + y)³ – 3xy(x + y)

⇒ x³ + y³ = (x + y)[(x + y)² – 3xy]

{Taking(x+y) Common}

⇒ x³+ y³= (x + y)[(x²+ y² + 2xy) – 3xy]

⇒ x³+ y³ = (x + y)(x² + y² – xy)

L.H.S= R.H.S

sin 82.5° = 0.9914

0.83694996final answer

Let theta = x

RHS:cosec x + cot x= 1/sin x + cos x/sin x= (1 + cos x) / sin= (1 + cos x)/sqrt(1 - cos^2 x) = sqrt(1 + cosx). sqrt(1 + cosx)/ sqrt(1 + cosx). sqrt(1 - cosx)= sqrt(1 + cosx)/sqrt(1 - cosx)= LHS

Hence proved

(2sin²x +sinx +1) = 2sin²x+2sinx-sinx-1= 2sinx(sinx+1)-1(sinx+1)= (2sinx-1)(sinx+1)

2sin²x-3sinx+1 = 2sin²x-2sinx-sinx+1=2sinx(sinx-1)-1(sinx-1)= (2sinx-1)(sinx-1)

so, the fraction becomes

limx→π/6 (2sinx-1)(sinx+1)/(2sinx-1)(sinx-1) =(sinx+1)/(sinx-1)

now putting x = π/6 we get the limit as

= (sin(π/6)+1)/(sin(π/6)-1) = (3/2)/(-1/2)= -3

If sec theta= 2cos = 1/2theta= 60°sin= √3/2tan = √34cos60°-√3sin60/tan60-cot60

= 2-3/2/√3-1/√3-1/2/2/√3-√3/4

Sin∅ = 1 / 3

So ,

Cosec∅ = 1 / sin∅

Cosec∅ = 3 .

Now ,

2Cot²∅ + 2

= 2 ( Cot² ∅ + 1 )

= 2 * Cosec²∅

[ By identity ]

= 2 * ( 3 ) ²

= 2 * 9

= 18

To proof:

tan²A - sin²A = tan² A sin²A

from LHS,

tan²A -sin²A

= (sin²A / cos²A) - sin²A……[tan A=sin A/cos A]

= (sin²A - sin²Acos²A) / cos²A

= sin²A (1- cos²A) / cos² A [tan A = sinA / cos A]

= tan²A sin²A= RHS

.•. hence proved

x² - y² - 9z² + 6yz

x² - ( y² + 9z² - 6yz)

x² - ( y - 3z)²

( x - ( y - 3z)) ( x + ( y - 3z))

( x - y + 3z) ( x + y - 3z)

define Poiseuilli's equation

Consider theta as A

cosA+sinA=√2cosAsquaring both the sides =>(cosA+sinA)^2=2cos^2A=>cos^2 A +sin^2 A+2sinAcosA=2cos^2 A=>cos^2 A-2cos^2 A+2sinAcosA= -sin^2 A=> -cos^2 A+2sinAcosA= -sin^2 A=> cos^2 A-2sinAcosA=sin^2 Aadding sin^2 A on both the sides => cos^2 A+sin^2 A-2sinAcosA=2sin^2 A=> (cosA-sinA)^2 =2sin^2 A

use square root both side,=> cosA-sinA=√2sinA

it's okay... but it's was to complement... I need more easy process to understand... but thanxxx for that

Taking L.H.S

consider (^) = power square

(sinA + secA)^ + (cosA + cosecA)^

(1/cosecA + secA)^ + ( 1/ secA. + cosecA)^

[( 1 + secA× cosecA)/cosecA]^ + [ (1 +cosecA×secA)/secA]^

(1 + secA×cosecA)^/cosecA^ + ( 1 + cosecA×secA)^/ secA^

taking common ,

(1 + secA× cosecA)^ (1/cosec^A + 1/ sec^A)

(1 + secA× cosecA)^ (sin^A +cos^A)

(1 + secA×cosecA)^ (1)

( 1 + secAcosecA)^

it has proved, L.H.S = R.H.S

thanks

cosx=3/5; sinx^2=(5)^2-(3)^2=25-9=16; sinx=4/5; sinx-cosx/2 tanx= 4/5 - 3/5/2 (4/5)= =4-3/5 /2(4/3)=-1/5/8/3= 1/5 × 3/8= 8x15/40=120/40= 120/40=12/4=3

cosx=3/5; sinx=4/5; tanx=4/3; (4/5)-(3/5)/2(4/3) = (4-3/5) x 3/8=(1/5)×(3/8)= 8+15/40=23/40

1/30 is the correct answer for this question

1/30 is correct answer

C.P.=11433013%of cost on repairs so total C.P.=1.13×114330=129,192.9now he has to make profit of 12%so S.P.=1.12×129,192.9=144,696.048

it's not correct

sorry it's correct

A trueaxiomcan not be refuted because the act of trying to refute it requires that very axiomas a premise. An attempt to contradict anaxiomcan only end in a contradiction. The term "axiom" has been abused in many different ways, so it isimportantto distinguish the proper definition from the others.

The following is a pretty good definition for any inverse function:

Letf(x)be a function that has an inverse and letf−1(x)be that inverse, thenf−1(f(x))=xandf(f−1(x))=x

Therefore, the inverse sine "undoes" what the sine does.

So,sin^-1(sin pi/6) = pi/6

sin^-1 and sin cancel with each other so the answer is pi/6

total money is x,also x be the 100%she spent 40% of total money for ingredients therefore40% of x=12

1/2*(1/6+1/5)=1/2*(11/30)=11/60Now 1/6=0.1671/5=0.2and 11/60= 0.188hence 0.188 lies in between 0.167 and 0.2 hence proved.

(1 - i) /(1 + i)

Multiply numerator and denominator by (1 - i)

i^2 = - 1

We get,(1 - i)*(1 - i) / (1 - (-1))= (1 - 1 - 2i)/2= - 2i/2= - i

Therefore real part is 0 and imaginary part is - 1

Solution is given below :

3x - 5y = -1 ....................... (eq. 1)

x - y = -1

⇒ y = x + 1 ........................(eq 2)

Substituting the value of y in eq. 1

⇒ 3x - 5(x + 1) = -1

⇒ 3x - 5x - 5 = -1

⇒ -2x = 4

⇒ x = -2

⇒ y = x + 1 = -2 + 1 = -1

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