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125u³ + 27 (5u)³ + 3³use a³ + b³ = (a + b) ( a² + b² - ab)(5u + 3) ( 25u² + 9 - 15u)

d=√(x2-x1)²+(y2-y1)²

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We have been given ,Sn = 4n -n^2

Just Think The Given Equation Of The Sum Of " n " Terms Is correct for all the terms in the AP , So If you put n=1 You get Sum of 1 terms which is actually the value of first term a1 .It goes like this , Sn = 4n - n^2 = 4(1) - (1)^2= 4 - 1 = 3

so The First Term Is 3.

Sum Of 2 Terms Is ...S2 = 4(2) - (2)^2 = 4

Sum of first 2 Terms means a1 + a2 , Now you know a1 and S2 , so S2 - a1 = a2 = 4-3= 1

Now the common difference , a2- a1 = d = 1-3 = -2.

Hence The Basic Method Of Doing That was this .

Solving for a10 , an , a3 we get ,a3 = -1 ;a10 = -15;an = 5 - 2n ;

Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2tan a+cot a=sin a/cos a +cos a/sin a=sin ^2 a +cos ^2 a/sin a cos a=1/(1/2)=2

sin²o + cos²o = 22 the answer is right

( sinx + cosx)^2=( V2)^2; ( sinx^2 + cosx^2 + 2sinxcosx) = 2; 2 sinxcosx = 2-1=1; sinxcosx = 1/2

sin θ + cos θ = √2

Squaring on both sides,(sin θ + cos θ)² = √2²

sin²θ + cos²θ + 2sinθ.cosθ = 2

1 + 2sinθ.cosθ = 2

2sinθ.cosθ = 2-1

2sinθ.cosθ = 1

sinθ.cosθ = ½

sinθ.cosθ = (1/√2) × (1/√2)

sinθ = cos θ = 1/√2

θ = 45*

1% = 1/100 15%(A+B) = 25%(A-B) 15(A+B)/100 = 25(A-B)/100 3(A+B) = 5(A-B) 3A+3B = 5A-5B 3B+5B = 5A-3A 2A = 8B A = 4B A = 4B*100/100 400% of B is equal to A

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Split the middle term as factors of 18. x^2 +6x+3x+18 =0x(x+6) +3(x+6) =0 (x+3)(x+6) =0 x=-3 x=-6

answer is 3/5 as sin theta is 3/4

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cosec sqr A - cot sqr A = 1 = (cosec A - cot A)*(cosec A + cot A) ....eq1

cosec A - cot A = p ..... eq 2 given

usin eq 1 we have cosec A + cot A = 1/p ----------- eq3

add eq2 and eq3 we get 2 cosec A = p + 1/ptherefore cosec A = 1/2 (p + 1/p)

First of all let us calculate the side of this equilateral triangle.

Let the sides of equilateral triangle be "s" cm.

Also, Area of equilateral triangle = √3/4×(s)²

Then, √3/4 × (s)² = 8√3

=> => s =√32m.

Now, by using the formula, we get :

Height of equilateral triangle

= √3/2 × √32

=

= cm.

breadth = 8 m.area = 120 sq mdiagonal = 17m

Let θ= x

cos x + sin x =√2 cos x

squaring on both side, we get......

cos2x + sin2x + 2cosxsinx = 2cos2x

2sinxcosx = 2cos2x - cos2x - sin2x

2sinxcosx = cos2x - sin2x

2sinxcosx = (cosx+sinx) (cosx - sinx)

2sinxcosx = (root2 cosx) (cosx - sinx)

2sinxcosx/root2 cosx = cosx - sinx

√2sinx = cosx - sinx

the correct answer is 0

cot x=3/4=b/ph=√3^2+4^2=√9+16=√25=5cosec x=h/p=5/4cos x=b/h=3/54×5/4+5×3/5=5+3=8

8 is the right answer

8 is the correct answer......

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The right answer is 8

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8 is the correct answer.......

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8 is the right answer

cosec x =5/4cos x=3/54*5/4+5*3/5=8

8 is the right ans.....

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8 is the correct answer

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ans. will be 8.

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the answer of this question will be 16

8 is the Right answer

8

in question perpendicular is 4 base is 3so by Pythagoras theorem hypotenuse is 5 hence in third quadrant cosec and cos is negative so answer is -8

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correct ans is 8 use pythogoras to theorem

8 is the right answer.

your questions is answer 15

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Answer is 32/3

since cot theta =b/p so first find h and put in cosec theta =h/p and cos theta =b/h and solve get answer 8

cot theta=3/4 so cosec theta= root over 1+9/16= 5/4 and sin theta = 4/5 so cos theta = root over 1-16/25=3/5 so 4cosec theta +5 cos theta =5+3=8

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4*5/4+5*5/3= 20/3 answer

(4*5/4)+(5*3/5)=8 is right

8 is tha right answer

so simple s right answer is 8.that mind solution.

so simple questionright answer is 8.that's mind solution.

value of 4 cosecx+5cosx=-8

cos45= 1/√2sec30= 2/√3cosec 30= 2now

1/√2/(2+2/√3)1/√2/(2√3+2)/√3

LHS =cosA/(1+sinA) +(1+sinA)/cosA

={cos²A +(1+sinA)²}/cosA.(1+sinA)

={cos²A+1+sin²A+2sinA}/cosA.(1+sinA)

use sin²∅ +cos²∅ =1

=(1+1+2sinA)/cosA(1+sinA)

=2(1+sinA)/cosA(1+sinA)

=2/cosA

=2secA = RHS

7x:5x i.e, length:breadth

perimeter=2(l+b)=2(12x)=24xi)l:b=7:5ii)b:l=5:7iii)l:p=7:24iv)b:p=5:24

(B) One

Explanation:The digits arranged in ascending order from left to right: 2345679Except 2, all others' position changes. Therefore, the number of digits which will retain the position is one.

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