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(sin5x-2sin3x+sinx)/(cos5x-cosx)=tanx

LHS

=[(sin5x-sin3x)-(sin3x-sinx)]/(cos5x-cosx)

=[(2cos4x.sinx)-(2cos2x.sinx)]/(-2sin3x.sin2x)

=2sinx.(cos4x-cos2x)/(-2sin3x.2sinx.cosx)

=(cos4x-cos2x)/(-2sin3x.cosx)

=(-2sin3x.sinx)/(-2sin3x,cosx)

=(sinx)/(cosx)

=tanx , Proved.

sin 5A + 2 sin 8A + sin 11A] / [sin 8A + 2 sin 11A + sin 14A]

= [ (sin 5A + sin 11 A) + 2 sin 8A] / [(sin 8A + sin 14A) + 2 sin 11A] (Rearranging)

= [ 2 sin (11A + 5A)/2 * sin (11A - 5A)/2 + 2 sin 8A] / [ 2 sin (14 A + 8A)/2 * sin (14A - 8A)/2 + 2 sin 11A]

{ Using sin C + sin D = 2 sin (C + D)/2 * sin (C - D)/2 }

= [ 2 sin 16A/2 * sin 6A/2 + 2 sin 8A] / [ 2 sin 22A/2 * sin 6A/2 + 2 sin 11A]

=2[ sin 8A * sin 3A + sin 8A] /2[ sin 11A * sin 3A + sin 11A] ( Taking 2 common in numerator and denominator)

= sin 8A[ sin 3A + 1]/ sin 11A[ sin 3A + 1]( Taking sin 8A common in numerator sin 11A common in denominarator)

= sin 8A / sin11A

2sin67.5cos22.5=2sin(90-22.5)cos22.5=2cos22.5cos22.5=1+cos45=1+1/root(2)

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Consider 252 and 324. here, a=324 and b=252

by euclid's division lemma-

a=bq+r, 0< or= r<b

324=252*1+72

252=72*3+36

72=36*2+0

therefore, HCF(252, 324)=36

Now consider 36 and 180. here a=180 and b=36.

by euclid's division lemma- a=bq+r, 0< or = r < b

180=36*5+0

therefore, HCF(180, 36)=36

Hence, HCF(180, 252, 324)=36

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bhai 3 ki power 2 h

please post your question clealry. power sign is different it is ^.

you can do the same just the n th term will be different

but 3^2 how to manage this

InΔ QTR∠ TQR +∠ QRT +∠ QTR = 180°⇒ 40° + y + 90° = 180°⇒ y = 180° - 130°⇒ y = 50°∠ QSP =∠ SPR +∠ SRPReason : Exterior angle = sum of interior angles⇒ x = 30° + y⇒ x = 30° + 50°⇒ x = 80°So, ∠x= 80° and∠ y = 50°

Tq

2 is the answer which req

5/(root(3)-root(5) multiply denominator and numerator with root(3)+root(5)=5*(root(3)+root(5))/(3-5)=-5/2(root(3)+root(5))

30-22=8=2^367-40=27=3^3416-200=216=6^3

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let the S.P = rs1so, sp=cpsp of 16 = 17rs as we know that, when do is greater there is profitPROFIT = selling price - cost price = 17-16=1 is profitgain%=gain/cp*100 1/16*1001/6.2gain %=6.2%

Ans :- 21b - 32 + 7b - 20b = b(21+7-20) -32 = b×8 - 32 = 8b - 32

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21b-20b+7b-3221b-13b-328b-32

The mode is the value which occur maximum number of times, that is, the mode has maximum frequency. If the maximum frequency occurs for more than 1 value, then the number of mode is more than 1 and is not unique.

Here it is given that the mode is 4. So,xmust be 4, otherwise it contradicts that the mode is 4. Hence,

x=4

12x³-17x²+26x-18 is the answer

12x³-17x²+26x-18 is the correct answer of the given question

x+42+68=180, x+110=180; x=180-110=70

x+48+68=180x=180-116x=64

16x2-24x+9-(40x2+20x+20x+10) / 8x2+4x-6x-3=3 (by cross multiplication)

16x2-24x+9-40x2-40x-10 / 8x2+4x-6x-3 = 3

-24x2-64x-1 / 8x2-2x-3 = 3

-24x2-64x-1=24x2-6x-9

48x2+58x-8=0

24x2+29x-4=0

Using factorisation,

24x2-3x+32x-4=0

(8x-1) =0 or (3x+4)=0

therefore, x= 1/8 or -4/3

only b) a) is not polynomial as its variable have negitive power.

give b statement also

your question is not visible

2A/3=(75/100)B=(6/10)C

2A/3=(3/4)B=(6/10)C

280A=90B=72C

A:B=9:8B:C=4:5

A:B:C=9:8:10

cotA= 7/8b/p= 7/8b/7=p/8= k(let)b= 7kp= 8kh^2=p^+b^2 = (8k)^2+(7k)^2 = 64j^2+49k^2 = 113k^2h= k√113sinA= p/h= 8k/k√113 = 8/√113(1-sinA)(1-sinA)=1-sin^2A{(a+b)(a-b)=a^2-b^2)=1-64/113=113-64/113= 49/113cos^2A= b^2/h^2 = 49k^2/113k^2=49/113(1+cosA)(1-cosA)=1-cos^2A= 1-49/113= 113-49/113= 64/113(!+sinA)(1-sinA)/(1+cosA)(1-cosA)= 49/113/64/113= 49/64

cotx=7/8; (hypo)^2=64+49=113; ( hypo)=V113; sinx=8/V113, cosx=7/V113; (1+8/V113)(1-8/V113)/ (1+7/V113)(1-7/V113)= (1+sinx)(1-sinx)/ (1+cosx)(1- cosx)= (1-sinx^2)/1-cosx^2)= (1-64/113)/(1-49/113)= (113-49/113) /(113-64/113)= (113-49)/(113-64)=64/49=8/7

cot¢= 7/8(i) (1+sin¢)(1-sin¢)/(1+cos¢)(1-cos¢)=(1-sin^2¢)/(1-cos^2¢)=cos^2¢/sin^2¢=cot^¢=(7/8)^2=49/64

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First of all cos 172 = -cos 8So now we have= cos 52 + cos 68 - cos 8

Applying Identity,cos A − cos B = −2 sin ½ (A + B) sin ½ (A − B)

let's look at our cos 52 - cos8, here let's A = 52, B=8cos 52 - cos 8 = -2sin[60/2]sin[44/2]cos52 -cos8 = - sin 22socos 52 + cos 68 + cos172= cos52 + cos68 - cos8= cos52 - cos8 + cos68= - sin22 + cos68but by the complementary relationshipsin 22 = cos68 , (since 22+68 = 90)so finally= -cos 68 + cos 68= 0

-cos x^2/sinx.-sinxcosx^2/sinx^2=cotx^2

numerater:cos(π+x)=-cos x.•. it lies in 3rd quadrant,where cosine is -ve. &, cos(-x) =-cos xSo,numerater:(-cos x)(-cos x)

Since the total number of outcomes is 7 and the chance of getting the number 5 is equally likely ,so the probability of getting 5 is equals to the number of desired outcome by total number of possible outcomes which is equals to 1 / 7.

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A non-leap year has 365 daysA year has 52 weeks. Hence there will be 52 thrusday for sure.52 weeks = 52 x 7 = 364 days .365– 364 = 1day extra.In a non-leap year there will be 52 Thursday and 1day will be left.This 1 day can beSunday, Monday,Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.Of these total 7 outcomes, the favourable outcomes are 1.Hence the probability of getting 53 Thursday= 1 / 7.∴ probability of getting 52 Thursday = 1 - 1/ 7 = 6 / 7.

There are 52 weeks in a way which means there will be 52 Sundays in a year.

There are 365 days in a year but we have 366 days in a leap year.

There are 7 days in a week.

If we multiply the weeks by the days we have 52x7 which equals to 364. This means that there are 2 extra days in a leap year which will make it 366.

The probability of having 52 Sundays in a leap year is thus: the remaining two days can be any of this formation:

Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, Saturday-Sunday.

However, to get 52 Sundays in a leap year, none of the remaining two days must be a Sunday.

Therefore, out of the 7 combinations above, that can be only realized 5 out of 7 times. The connection "Sunday-Monday and Saturday-Sunday" most be scraped off.

The probability of having 52 Sundays in a leap year is therefore 5/7.

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