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LHS=sinx/cosx+cosx/sinxTaking LCM=sin2x+cos2x/sinxcosx=1/sinx cosxRHS=1/cosx *1/sinx=1/sinxcosxLHS=RHSHence,Proved.

pahchan ke Vrat Mein Sone Ka batao

-16cotx+25tanx-81x+c

-16cotx+25tanx-81x+c is the correct answer h

opan square and arrange the equation after some trims is become one and some trims solve easy

41x +16cotx - 25tanx+ 40

-16/tan2.-81x+25tan(X)+C is the correct answer by PREM KUMAR SINGH

16cotx+25tanx-81x+c ans hoga

bhag Teri to

CotAcosA/cotA+cosA=[(cosA/sinA)cosA]/[(cosA/sinA)+cosA]=(cos²A/sinA)/[(cosA+sinAcosA)/sinA]=cos²A/cosA(1+sinA)=cosA/(1+sinA)=[cosA(1-sinA)]/[(1+sinA)(1-sinA)][multiplying the numerator and the denominator with (1-sinA)]=[cosA(1-sinA)]/(1-sin²A)=(cosA-cosAsinA)/cos²A=[(cosA-cosAsinA)/sinA]/(cos²A/sinA)[dividing the numerator and the denominator by sinA]=[(cosA/sinA)-(cosAsinA/sinA)]/(cosA/sinA)cosA=(cotA-cosA)/cotAcosA (Proved)

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what sonam gandhi has done is that she solved out this sum by taking out the formul of cot a and cos a.

And then she had proceeded the sum line by line or you can say that step by step

with the help of the property of logarithms

u can find more from this method

To prove: cubic root of 6 is irrationalSolution:Assume cube root 6 is rational. Then let cube root 6 = a/b here a and b are co prime numbers Cubing both sides : 6=a^3/b^3

a^3 = 6 b^3 a^3 = 2(3 b^3) Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number divides the product of two integers then it must divide one of the two integers Since all the terms here are the same we conclude that 2 divides a. Now there exists an integer k such that a=2k Substituting 2k in the above equation 8k^3 = 6b^3 b^3 = 2{(2k^3) / 3)} Here 2 divides b^3. thus 2 divides b. Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong.

diffrentiation xunder root x

=sinA/1+cosA + 1+cosA/sinA = [sin^2A + (1 + cosA)^2]/sinA(1 + cosA)= (sin^2A + 1 + cos^2A + 2cosA)/sinA(1 + cosA) = (1 + 1 + 2cosA)/sinA(1 + cosA) [since,sin^2A + cos^2A = 1] = (2 + 2cosA)/sinA(1 + cosA) = 2(1 + cosA)/sinA(1 + cosA) = 2/sinA = 2 cosecA [since 1/sinA = cosecA] = R.H.S

y= cot^1-(V cosx)- tan^1-( Vcosx) let cot^1-(Vcosx)=p; sinQ=Vcosx/ V(1+ cosx); y=p-q siny= sin(p-q)=sinpcosq- cops sinq = 1/V( 1+cosx).1/V1+ cosx - Vcosx/V1+cosx. V cosx/V( 1+ cosx)=(1- cosx)/(1+ cosx) = 2 sinx^2/2/2 cosx^2/2= tanx/2^2; siny= tan^2(x/2)

index form = 30^1/7

is the answer

sin4A/sin5A=tan4A = sin4A / cos4Asin5A = cos4A

=> sin5A = cos(π/2 - 5A) = cos4Aπ/2 - 5A = 4A9A = π/2A = π/18

wrong hai

wrong answer

real answer is 66•x power 4 h

1/number of days require = 1/12+1/15 = (5+4)/60 = 9/60 = 3/20 so number of days required = 20/3

r u sure about answer and thanks for the solution

use this

cost of 12 cans = 840one can's cost = 840/12= 703 dozen= 3*12=36 cans70*36= 2520 rupee

thank u sir

tan5A= cot2Atan5A= tan(pi/2-2A)general solution is5A=npi+(pi/2-2A)7A=(2n+1)pi/2

answer is 15°

thanku

5) 20xy-5xy=15xy6)x^2 +xy =x(x+y)

Suppose we consider ,√6 is a rational number .

Then we can express it in the form of a/b

∴√6=a/b, where a and b are positive integer and they are co-prime ,i.e.HCF(a,b)=1

∴√6=a/b

=>b√6=a

=>(b√6)²=a² [squaring both sides]

=>6b²=a²………..(1)

here,a² is divided by 6

∴a is also divided by 6. [we know that if p divides

a²,then p divides a]

∴6|a

=>a=6c [c∈ℤ]

=>a²=(6c)²

=>6b²=36c² [from (1)]

=>b²=6c²

here,b² is divided by 6,

∴b is also divided by 6.

∴6|a and 6|b

we observe that a and b have at least 6 as a common factor .But this contradicts that “a and b are co-prime .”

It means that our consideration of “√6 is a rational number” is not true.

Hence,√6 is a irrational number.

listen mister a rational number is of a fixed value I.e √6 gives a value which goes on. Thus √6 cannot be called as rational number