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Multiplicative identity is a term which when multiplied by any number gives the same number.definition ; c is multiplicative identity ifc×b = b, b be any real number.1 is a multiplicative identity

Given,4 tan = 3∴ tan = 3/4

As we know,Tan= Perpendicular / baseTan= 3/4Now,Hypotenuse =√3² + 4² =√9+16 =√25 = 5

Using this, we can find:Sin = Perpendicular / Hypotenuse = 3/5Cos = Base / Hypotenuse = 4/5

ATQ,(4sin - cos + 1)/(4sin + cos - 1)= (4 × 3/5 - 4/5 + 1)/(4 × 3/5 + 4/5 - 1) [putting values]= (12 - 4 + 5)/(12 + 4 - 5)= 13/11

cos theeta= 4/5sin theeta= ✓(1-16/25= 3/5sec theta= 5/4cot theeta= cos/sin= 4/5/3/5= 4/3(1-sec^2 theeta)= (1-25/16)[ -9/165/4*(4/3)/(-9/16)(5/3)*(-16/9)= -80/27thanksplease like the solution 👍 ✔️

( a sin - b cos )² + ( a cos + b sin)²

= a² ( sin² + cos²) + b² ( sin² + cos²) - 2ab sin cos + 2 ab sin cos

= a² + b²

sin4x+ sin2x= sim2x+2sinxcosx= sin2x(1+2cos2x)=sin4x=2sinxcosx for the denomater 1+ cos2x+ cos4x; using cos2theta= cos^(2)theta- sin^(2)theta= 2 cos^(2)theta-2; we substitute cos4x as 2cos^(2)2x -1, so, 1cos2x+ cps4x=1+ cos2x + 2cos^(2)2x-1= cos3x+2 cos^(2)2x, I,e. cos2x(1+2cos2x), so the numerator and demonator now have a common factor as (1+2 cos2x), so dividing we sin2x/ cos2x left which is equivalent to

which is equivalent to tan2x

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thnks mam

5x^2 -6x -2 = 0

x^2 -(6/5)x -(2/5) = 0

(x)^2 - [2.(3/5).x] + (3/5)^2 - (3/5)^2 -(2/5) = 0

[x-(3/5)]^2 = (2/5)+(9/25)

[x-(3/5)]^2 = (19/25)

x-(3/5) = √19/5

x = (3/5) ± √19/5

Roots of the equation are

x = [3+(√19)]/5 & x = [3-(√19)]/5

thanks

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b-[b-(a+b)-{b-(b-a-b)+2a

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7√3/(√10+√3)-2√5/(√6+√5)-3√2/(√15+3√2) isme sayugmi se multiple karne par7√3/(√10+√3)×(√10-√3)/(√10-√3)-2√5/(√6+√5)×(√6-√5)/(√6-√5)-3√2/(√15+3√2)×(√15-3√2)/

(√15-3√2)=7√3(√10-√3)/7-2√5(√6-√5)/1-3√2(√15-3√2)/-3=√30-3-2√30-10+√30-6=-19

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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 HENCE PROVED

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Here s cannot be 40 It should be 30.then solving that we get 479.06.

lim x ---> 0 x sec(x)

lim x ---> 0 { 0× 1} = 0

1/(secx-tanx) - 1/cosx= secx + tanx -secx= secx - (secx-tanx)= 1/cosx - (sec²x-tan²x)/(secx+tanx)=1/cosx - 1/(secx+tanx)

Hi,

Here is the answer to your question.

GIVEN : The perimeter of a triangular field = 540m

Let the sides are 25x , 17x , 12 x

Perimeter of a ∆ = sum of three sides

25x + 17x + 12x = 540

54x = 540

x = 10

1st side (a) - 25x = 25×10= 250m

2nd side(b)= 17x = 17×10= 170m

3rd side (c)= 12x = 12 × 10 =120m

Semi - perimeter ( S) = a+b+c/2

= (250 + 170+120)/2 = 540/2 = 270 m

Area of the ∆= √ S(S - a)(S - b)(S - c)

[By Heron’s Formula]

= √ S(S - 250)(S - 170)(S - 120)

= √ 270(270 - 250)(270 - 170)(270 - 120)

= √ 270× 20×100×150

= √ 81000000

Area of the ∆= 9000 m²

Cost of ploughing the field = (10/18.80) ×9000 = 4787 rupees

Let the money recieved by kiran be x

then the money recieved by jennifer will be (x + 500)

and the money received by salma will be (x + 1000)

We kbow that their sum =13500

so, x+(x+500)+(x+1000) = 13500

x+x+x+500+1000 = 13500

3x+1500 = 13500

3x = 13500-1500

x = 12000/3

x = 4000

So, the money recieved by kiran is 4000

Si, the money receibed by jennifer will be 4000+500 =4500

total money = 13500let kiran's money be xsalma's money = x +1000jenifer's money= x + 500an equation formedx +(x + 1000) +(x + 500) = 135003x + 1500 =135003x=13500-1500x=12000/3x=4000kiran's money=4000jenifer money= x + 500 = 4000+500 =4500

total money =13500salma money==•^=€÷Π¥=•¶|×^

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CGS unit for length is cmhence area will be cm^2

its integration dear

Sec(45+a)sec(45-a)

=> 1/cos(45 + a)*cos(45 -a)

=> 1/(cos45cosa – sin45sina) (cos45cosa+sin45sina)

=> 2/(cosa -sina) (cosa + sina)

=> 2/(cos^2a -sin^2a)

=> 2/cos2a

=> 2sec2a

15 is the correct answer

tnx bro