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(1-cos2A)/(sin2A)=2sin^2A/2sinAcosA=sinA/cosA=tanA

1980if there is anymultiple by any 0digitquestion like17845×100=1784500

d. 198X10=1980 is the Answer

1)198×10=1,9802)305×100=30,500plz like thanks

After simplification ,answer will be= 0.06

answer kese aya

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Just subtract the remainder from the dividend:121- 1 = 120183 -3 = 180305 -5 = 300

The required number is the HCF of 120, 180 and 300...which is 60

hence the required number is 60

Thanks

The lac operon of E. coli contains genes involved in lactose metabolism. It's expressed only when lactose is present and glucose is absent.Two regulators turn the operon "on" and "off" in response to lactose and glucose levels: the lac repressor and catabolite activator protein (CAP). The lac repressor acts as a lactose sensor. It normally blocks transcription of the operon, but stops acting as a repressor when lactose is present. The lac repressor senses lactose indirectly, through its isomer allolactose.

sinA = bsinB....(1)

tanA= atanB

(sinA/cosA) = a(sinB/cosB)....(2)

substuting sinB value from equation (1)

cosB = (a/b) cosA......(3)

sin2A = b2sin2B

1-cos2A = b2(1-cos2B)

substituting equation (3)

1-cos2A = b2[1 – ((a2/b2)cos2A)]

1 – cos2A = b2– a2cos2A

a2cos2A – cos2A = b2– 1

cos2A = (b2– 1)/(a2– 1)

Distance measured to the left of the mirror is -ve.

Distance measured to the right of the mirror is +ve.

cosA(cosec A -1) LHS=….............................. cosA(cosec A + 1) =cose A- 1/cosecA + 1 RHS

LHS :

(secA + tanA)²

( 1/cosA + sinA/cosA)²

(1 + sinA)² / cos²A

we know that ,

cos²A =1 - sin²A

(1+sinA) (1 + sinA) / {1 - sin²A}

Also ,

1 - sin²A =(1+sinA) (1 - sinA) [∵a² -b² = (a+b) (a - b) ]

(1+sinA) (1 + sinA) /(1+sinA) (1 - sinA)

(1 + sinA )gets cancelled on numerator and denominator ,

=> 1 + sinA / 1 - sinA

sinA = 1/cosecA

=> 1 + ( 1/cosecA) / [ 1 - (1/cosecA) ]

=>cosecA+1/cosecA÷cosecA-1/cosecA

=> cosec A gets cancelled ,on both sides ,

hence ,

LHS becomes ,

cosecA + 1/cosecA - 1

(y+z)(x^2+xz+yz+xy-yz)-x(xy+y^2-yz)+x(yz-z^2-xz)=x^2y+x^2z+xyz+xz^2+xy^2+xyz-x^2y-xy^2+xyz+xyz-xz^2-x^2z=4xyz

b=1/25. is the right answer

Let theta = x

LHS: (cotx+cosecx)-1/(cotx-cosecx+1)

=(cosecx+cotx)-(cosec²x-cot²x)/(cotx-cosecx+1)

= (cosecx+cotx)[1-(cosecx-cotx)]/(cotx-cosecx+1)

= (cosecx+cotx)(cotx-cosecx+1)/(cotx-cosecx+1)

= (cosecx+cotx)

= RHS

Hence proved

(2/9)^3 +( 2/9)^-6=(2/9)^2m-1(2/9)^-3=(2/9)^2m-12m-1=-32m=-2m=-1

thanks

Cosec A + cot A - 1 / cot A - cosec A + 1we know that,cosec ² A - cot ² A = 1substituting this in the numeratorcosec A + cot A -(cosec ² A - cot ² A) / (cot A - cosec A + 1)x²-y²= (x+y)(x-y)cosec A + cot A - (cosec A + cot A) (cosec A - cot A) / (cot A - cosec A + 1)taking common(cosec A + cot A)(1-cosec A + cot A) / (cot A - cosec A + 1)cancelling like terms in numerator and denominatorwe are left with cosec A + cot A= 1/sin A + cos A/sin A= (1+cos A) / sin A

not understood plzz explain it clearly

thanku sister....

3x^2-x-4= 0

Here is your answer

=3x²-x-4=3x²-(4-3)x-4=3x²-4x+3x-4=x(3x-4)+1(3x-4)=(x+1)(3x-4)

x+1=0x=-1

OR

3x-4=03x=4x=4/3

alpha + beeta= 1/3

3x^2 + 5y - zGiven x = y + z, y =1/2zX = 3/2zThen 3(3/2z)^2 + 5/2z +z27/4z ^2 + 7/2zValue of z = - 4Put value of z in given equation 27/4 (16) - 1427*4 - 14108 - 1484 ans

Given,

x + y = w + z

To Prove,

AOB is a line or

x + y = 180° (linear pair.)

Proof:

A.T.Q

x+ y + w + z = 360° (Angles around a point.)

(x + y) + (w + z) = 360°

(x + y) + (x + y) = 360°

(Given x + y = w + z)

2(x + y) = 360°

(x + y) = 180°

Hence, x + y makes a linear pair.

Therefore, AOB is a straight line.

x+y+z+w=360now, x+y=z+w=180so, AOB is a line

Angle EHK = angle HKLangle EHK= (180-60 )+25thereforeangle HKL= 120° + 25° =145°

vi) LHS = √(1 + sin∅)/√(1 - sin∅)

= √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅)

= √(1 + sin∅)²/√(1 -sin²∅)

= (1 + sin∅)/√cos²∅

= (1 + sin∅)/cos∅

= 1/cos∅ + sin∅/cos∅

= sec∅ + tan∅ = RHS

vii) let theta =x

=sinx(1 - 2sin2x)/ cosx(2cos2x - 1)

=sinx[1 -2(1 -cos2x)]/cosx(2cos2x - 1)

=sinx[1 - 2 + 2cos2x]/cosx(2cos2x - 1)

=sinx[2cos2x - 1]/cosx(2cos2x - 1)

=sinx/cosx

= tanx = rhs

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take the denominator of lhs to rhs, hence proved

explain in easily method

x+ 1/x=V5; ( x + 1 / x)^2= x^2+1/x^2+2x(1/x) = x^2+1/x^2+2,; ( x+1/x)^2-2= x^2 + 1/ x^2

( x^2+1/x^2)=(x+1/x)(x+1/x)=V5 x V5=5

3 is right answer of your question

Let √5+√5+√5...=Xhencex=√5+xsquaring both sideshencex^2=5+Xx^2-x-5=0x=1+-√1+20/2x=1+√21/2 and x=1-√21/2

thank you for sending

thank you for sending my answer

Let x = daily wage

a1 = No. of days he is expected to work

a2 = No. of days he actually worked

Wages for the No. of days he was expected to work =

So,

a1x = 495.60----- (i)

Wages for the days he worked = 389.40

So,

a2x = 389.40 ----- (ii)

From (i)

x = 495.60/a1

Substitute this in (ii)

a2(495.60/a1) = 389.40

a2/a1 = 78/100

From this we can see that he was supposed to work for 100 days but he worked for only 78days.

The right answer is -11/28

the correct answer is-11/28

thanks for raising the doubt with us

6kg= 6000gnow ratio6000/40= 600/4= 300/2= 150:1