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Ans :- Validating Statements (&) If p and q are mathematical statements, then in order to show that the statement “p and q” is true, the following steps are followed. Validating Statements (or) Validating Statements (if -then) Validating Statements (if and only if) Validate by contradiction.

2sin²x-1/cosx*sinx = sinx/cosx - cosx/sinx

2sin²x-1/cosx*sinx = sin²x-cos²x/sinx*cosx

2sin²x-cos²x-sin²x = sin²x-cos²x

= 0

Hence proved

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

divide numerator and denominator with

sinA , we get

= ( cotA - 1 + cosecA)/( cotA+1 - cosecA )

= [cotA+cosecA- 1 ]/[ 1 - cosecA + cotA ]

=[(cotA+cosecA)-(cosec²A-cot²A)]/(1-cosecA+cotA)

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

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

= cosecA + cotA

= RHS

i) Multiplying by (sin²B)(cos²B), the given expression changes as:

(cos⁴A)(cos²B) + (sin⁴A)(sin²B) = (sin²B)(cos²B) -------- (1)

ii) (cos⁴A)(cos²B) = {(1 - sin²A)²}*(1 - sin²B)

= (1 - 2sin²A + sin⁴A)(1 - sin²B)

= 1 - 2sin²A + sin⁴A - sin²B + 2sin²A*sin²B - sin⁴A*sin²B

iii) Substituting this in (1) above,

1 - 2sin²A + sin⁴A - sin²B + 2sin²A*sin²B - sin⁴A*sin²B + sin⁴A*sin²B = sin²B*cos²B

==> 1 - 2sin²A + sin⁴A - sin²B + 2sin²A*sin²B = sin²B(1 - sin²B) = sin²B - sin⁴B

Rearranging, 1 - 2(sin²A + sin²B) + (sin⁴A + 2sin²A*sin²B + sin⁴B) = 0

==> 1 - 2(sin²A + sin²B) + (sin²A + sin²B)² = 0[This is in the form of a² - 2a + 1, where a = (sin²A + sin²B)]

==> {(sin²A + sin²B) - 1}² = 0

==> sin²A + sin²B = 1 ---- (2)

iv) From (2) above, sin²A = 1 - sin²B; ==> sin²A = cos²B ------- (3)Similarly, sin²B = 1 - sin²A; ==> sin²B = cos²A -------- (4)

v) So, sin⁴B/cos²A = sin⁴B/sin²B [Substituting from (4) above]==> sin⁴B/cos²A = sin²B -------- (5)

cos⁴B/sin²A = cos⁴B/cos²B [Substituting from (3) above]==> cos⁴B/sin²A = cos²B -------- (6)

Adding (5) & (6),

(sin⁴B/cos²A) + (cos⁴B/sin²A) = sin²B + sin²B = 1

Thus it is proved that, (sin⁴B/cos²A) + (cos⁴B/sin²A) = 1

c= 2πr145.2= 2×3.14×rr= 23.12md= 42.24m

Let theta = x

LHS:cos x + sin x/ sin x

= cos x/sin x + sin x/sin x

= cot x + 1

= 1 + cot x

= RHS

Hence proved

RHS√1-cosA/1+cosAby 1-cosA multiply in numerator and denominator√(1-cosA)(1-cosA)/(1+cosA)(1-cosA)=√(1-cosA)^2/1-cos^2A= √(1-cosA)^2/sin^2A=1-cosA/sinA=1/sinA - cosA/sinA = cosecA - cotA =LHSproved

cos^-1(2x-1)

Let x= cos^2A , or cosA=+/-x

=cos^-1(2cos^2A-1)

=cos^-1.cos2A

=2A

=2.cos^-1(x)^1/2

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Vibrationis a mechanical phenomenon wherebyoscillationsoccur about an equilibrium point.Oscillationis the repetitive variation, typically in time, of some measure about a central value (often a point of equilibrium) or between two or more different states. The termvibrationis precisely used to describe mechanicaloscillation.Vibrationcan be desirable:for example, the motion of a tuning fork, the reed in a woodwind instrument or harmonica, a mobile phone, or the cone of a loudspeaker.Familiar examples of oscillation include a swingingpendulumandalternating current.

Circle dividing the plane: Circle divides the plane on which it lies intothreeparts. They are:

inside the circle, which is also called the interior of the circle;

the circle and

Outside the circle, this is also called the exterior of the circle.

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No, It's is not possible as sum of all angles are 180 ° but here it's 190°. So no ☺️👌.nice question

Let theta = xLHS:sinx(1 + tanx) + cosx(1 + cotx)= sinx(cosx + sinx)/cosx + cosx(sinx + cosx)/sinx= (sinx + cosx)[sinx/cosx + cosx/sinx]=(sinx + cosx) [(sin^2 x + cos^2 x) /sinx. cosx]= (sinx + cosx) /sinx. cosx= 1/cosx + 1/sinx= secx + cosecx= RHS

Hence proved

thanks

answer

LCM × HCF = Product of numbersProduct of numbers = xy

xy should be the answer

1 upon 16 cancel 1 upon 2

1/2 is right answer hai

1/2 is the correct answer

1/2 is the correct answer of this question

1/2 answer your question

sinx/cosecx + cosx/secx

(sinx)^2+(cosx)^2

=1

here is a solution for a similar question. hope it helps

Since an x² is produced, we know that the factorised version will be (x+a)(x+b). To find a and bwe need to consider the relationship between 2 and -8. It is going to be factorised into two brackets, due to the x², so we need to break the 2 down into 2 numbers that will add to give 2 and multiply together to give -8. The 8 is negative so one of these numbers must also be negative. These 2 numbers must be 4 and -2. Then substitude these numbers in to be a and b so the equation factorised is (x+4)(x-2)

Concept :- If (a, b) is the point on the circle and (h,k) is the centre of circle, then radius (r) = √{(a-h)² + (b-k)²}. Then use equation of circle is ( x - h)² + (y - k)² = r²

Here, centre = (2, 2) and point on the circle is (4, 5) Then, radius (r) = √{(4-2)² +(5-2)²} r = √13

Now, equation of circle is (x - h)² + (y - k)² = r²(x - 2)² + (y - 2)² = √13²x² + 4 - 4x + y² + 4 - 4y = 13 x² + y² - 4x - 4y -5 = 0

Hence, equation of circle is x² + y² - 4x - 4y -5 = 0

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Sum of n terms of geometric series 1 + 2/3 + 4/9+.....

Sn = a(1-r^n)/(1-r)For given series a = 1, r = 2/3

Then,Sn = 1(1-(2/3)^n)/(1-2/3) = 3(1 - (2/3)^n)

Sum of first 5 termsS5 = 3(1 - (2/3)^5) = 3(1 - 32/243) = 3(243-32)/243 = 211/81

Thanks

p(x)= x^2-2x-15sox^2-(5-3)x-15= 0x^2-5x+3x-15= 0x(x-5)+3(x-5)x= -3x= 5

alpha+ beeta= 5-3= 2alpha*beeta= -15

the answer is forty

1+4=(4+1)=5, 2+5=(2×5=10+2)=12, 8+11=88+8=96

1+4=(4+1)=5, 2+5=(2×5=10+2)=12, 8+11=88+8=96 answer

8+11=19..........................💥

1+4=5 2+5=12 3+6=21 8+11=19

the answer is 96 nani valley is right

8 + 11 = 11x8=88+8=96

96 is the correct answer

4×1=4+1=55×2=10+2=126×3=18+3=2111×8=88+8=96

96 is the my answer to this question

1+4=55+2+5=1212+3+6=2121+8+11=40

40 is the correct answer

96 is right answer

if 4×1+1=52×5+2=123×6+3=21then 8×11+8= 96

4*1+1=55*2+2=126*3+3=217*4+4=328*5+5=449*6+6=6010*7+7=7711*8+8=96👍

40 is right answer..........

sorry 96 is right answer

40 is the right answer

etna phaltu qution ha ya

40 is right the correct answer

x/2- 2x/2-4/6=5/3-2x/2-4/6=5/3-2x/3=14/6x=-7/2

answer is 3.

please post it by writing it clealry

iii)

3x - y = 3 ...(1)

9x - 3y = 9 ....(2)

As observe (1) × 3 = (2)

Therefore , there are infinite solution

Hence proved