Explore topic-wise InterviewSolutions in .

1. -cos2x/22. sin3x/33. e^2x/24. (ax+b)^3/3a5. -cos2x/2-4/3e^3x

Since 56 and 34 are complementary angles

cos56°/sin34°= sin (90-56)/sin 34= sin 34°/sin34°= 1

a=34 d=-2 tn=a+(n-1)d 10=34+(n-1)-2 -14=(n-1)(-2) n-1=7 n=8 Sn=n(2a+(n-1)d)/2 S8 =8(2×34+(8-1)(-2))/2S8=8(68-14)/2S8=4(54) S8 = 216

LHS : apply SIN A+ SIN B and COS A + COS B formulae

2sin4xcosx / 2cos4xcosx ( 2 and cos x get cancelled )

we are left with : sin4x / cos 4x = tan4x ( since , sinx / cosx = tanx )

comment below if any doubt

1 + sin^3x + cos^3x =3/2*sin2x

1+(sinx+cosx)(sin^2x-sinxcosx+cos^2x)=3sinxcosx

1+(sinx+cox)(1-sinxcosx)=3sinxcosx

Let s=sinx+cosx

s²=1+2sinxcosx

1+s(1-(1-s²/2))=3(1-s²/2)

s^3+3s^2-3s-5=0

(s+1)(s^2+2s-5)=0

range of s is [-√2,√2]

s=-1

s=√2(sin(x+π/4))=-1

(sin(x+π/4))=-1/√2

x+π/4=-π/4+n(-1)^nπ

3x + x power 2 ggyyyyhhuhhhjjjjjjjiiiiiiiiiooiiiiiiii

x²+2x+1factors(x+1)(x+1)is the best answerx+1)

i) let (X+3) be m m²-8m+16m²-4m-4m+16m(m-4)-4(m-4)(m-4)(m-4)(m-4)²replace m by (X+3)((X+3)-4)²(X+3-4)²

a)2/10b)7/10c) 25/100=1/4d)245/1000e)301/100f)38/10g) 503/100

(a) ans=Answer:

0.2 written as a fraction is210which simplifies to15

Explanation:

When changing a decimal to a fraction, think of the decimal "out of 1." In mathematical terms, "out of" generally suggests division.

For example on this question, think of 0.2 as 0.2 out of 1, which would be shown as0.21

To more easily simplify this, multiply both the top and bottom by 10:

0.21⋅1010

You are able to do this because1010is equal to1and multiplying by a number and1results in the same number.

Then you get210. Both2and10are divisible by2:

22=1and102=5

So the final answer is15.

h)375/100i) 83.25/100j)4356/100

2/10. 7/10. 25/100. 245/10000. 301/100 . 38/10. 503/100. 375/100. 8325/100

please mark it as best

2/107/1025/100245/1000310/10038/100503/100375/1008325/1004556/100

2/10 . ... 7/1025/100245/1000310/10038/100503/100375/1008325/1004556/100

a) 0.2 = 2/10 = 1/5b) 0.7 = 7/10c) 0.25 = 25/100 = 1/4d) 0.245 = 245/1000e) 3.01 = 301/100f) 3.8 = 38/10g) 5.03 = 503/100h) 3.75 = 375/100i) 83.25 = 8325/100j) 45.56 = 4556/100

Meena 's position :- sBeena is 8 step aheadBeena ' s position :- s+8Leena is 7 step behindLeena ' s position :- s-7Total number of steps :- 4s-10

The result should be : If A + B + C = π, then

cos 2A + cos 2B + cos 2C = -1 - 4 cos A cos B cos C._______________________________

... LHS

= ( cos 2A + cos 2B ) + cos 2C

= [ 2 cos (A+B) cos (A-B) ] + cos 2C

= [ 2 cos (π-C) cos (A-B) ] + ( 2 cos² C - 1 )

= [ 2( -cos C ) cos (A-B) ] + 2 cos² C - 1

= -1 - 2 cos C cos (A-B) + 2 cos² C

= -1 - 2 cos C [ cos (A-B) - cos C ]

= -1 - 2 cos C [ cos (A-B) - ( -cos (A+B)) ]

= -1 - 2 cos C [ cos (A+B) + cos (A-B) ]

= -1 - 2 cos C [ 2 cos A cos B ]

= -1 - 4 cos A cos B cos C

= RHS ........................................... Q.E.D.

Cos⁴π/8+cos⁴3π/8+cos⁴5π/8+cos⁴7π/8

=cos⁴π/8+cos⁴3π/8+{cos(π/2+π/8)}⁴+{cos(π/2+3π/8)}⁴

=cos⁴π/8++cos⁴3π/8+(-sinπ/8)⁴+(-sin3π/8)⁴

=sin⁴π/8+cos⁴π/8+sin⁴3π/8+cos⁴3π/8

={(sin²π/8)²+(cos²π/8)²}+{(sin²3π/8)²+(cos²3π/8)²}

={(sin²π/8+cos²π/8)²-2sin²π/8cos²π/8}+{(sin²3π/8+cos²3π/8)²- 2sin²3π/8cos²3π/8}

=1-2sin²π/8cos²π/8+1-2sin²3π/8cos²3π²/8

=(1/2){4-(2sinπ/8cosπ/8)²-(2sin3π/8cos3π/8)²}

=(1/2)[4-(sinπ/4)²-(sin3π/4)²][∵, sin2A=2sinAcosA]

=(1/2)[4-(1/√2)²-(cosπ/4)²][∵, sin3π/4=sin{(π/2×1)+π/4}=cosπ/4]

=(1/2)[4-1/2-(1/√2)²]

=(1/2)[4-(1/2+1/2)]

=(1/2)(4-1)

=3/2 (Proved)

a³-b³=(a-b)³+3ab(a-b)1+tan²x=sec²x(sec²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x) =1³+3sec²xtan2x(1) =1+3tan²xsex²xhence proved

Sol-Given: Sin 10 Sin 30 Sin 50 sin 70⇒ (1/ 2) Sin 10 Sin 50 sin 70 multiply and divide with 2 Cos 10

⇒2 Cos 10 Sin 10 Sin 50 sin 70/ (2 x 2 Cos 10) { 2SinA Cos A = Sin2A }

⇒ Sin 20 Sin 50 sin 70 /(4 Cos 10)multiply and divide with 2

⇒ 2 x Sin 20 Sin 50 sin (90 - 20) /(2 x 4 Cos 10)

⇒ 2 Sin 20 Cos 20 Sin 50 / 8Cos 10

⇒ Sin 40 Sin 50 / 8Cos 10{ 2SinA Cos A = Sin2A }

Againmultiply and divide with 2.

⇒ 2 x Sin 40 Sin ( 90 - 40) / 2 x 8Cos 10.

⇒ 2 x Sin 40 Cos 40 / 2 x 8Cos ( 90 - 80)

⇒ Sin 80 / 16Sin 80 = 1 / 16.

ii baala karo solve

thanx aap mujhe hamesha answer dete ho

cos3A=4cos³A-3cosAA=30

Cos90=4cos³30-3cos30

0=4(√3/2)³-3(√3/2)0=4(3√3/8)-3√3/20=3√3)2-3√3/20=0

Verified

in the first expressioncos^2π/2=00-15*1*1/√2-4*√3/2*1/√2*1/2-15/√2-√3/√2-15-√3/√2

don't understand it

by replacing y with -ycos(x-(-y))=cosxcos(-y)+sinxsin(-y)cos(x+y)=cosxcosy-sinxsiny

cos30=√3/2and cos3*30=cos90=0hence4cos^330-3cos30=4*3√3/8-√3/2=√3/2-√3/2=0LHS=RHS

thanks

Compare x² + 4x + 2a with ax² +bx + c ,

a = 1 , b = 4 , c = 2a

According to the problem given ,

α and α / 2 are the zeroes of the polynomial ,

1 ) sum of the zeroes = - b / a

α + α/ 2 = - 4 / 1

( 2α + α ) / 2 = 4

3α = 8

α = 8 / 3

α² = 64 / 9 -----( 1 )

Product of the zeroes = c / a

α × α / 2 = 2a / 1

α² = 4a -------( 2 )

Therefore ,

Equation ( 1 ) = equation ( 2 )

64 / 9 = 4a

64 / ( 9 × 4 ) = a

16 / 9 = a

a = 16 / 9

I hope this helps you.

Number of parking spaces = 200Truck parking spaces =(10/100)*200 = 20Car parking spaces = (75/100)*200 = 150

Motorcycle parking spaces= 200 - (150 +20)= 200 - 170 = 30

Ratio of his saving to his pay720/45000.16Ratio of his expenditure to his pay(4500-720)/45003780/45000.84

Total investment is in the ratio of 5:7so A 's part will be5/12 of totaland B part will be 7/12 part of totalso if 7/12 part is 1370if 7/12 or (58.33) percentage is 1370then total profit (or 100 percentage)will be (1370/58.33*100=2348.70)

1/(sec x - tan x) - 1/cos x = 1/cos x - 1/(sec x + tan x)

1/sec x - tanx + 1/sec x + tan x= 2/cos x

LHS:1/sec x - tanx + 1/sec x + tan x=sec x + tanx + sec x - tan x/ sec ^2 c + tan^2 x= 2sec x /1= 2/cos x= RHS

Hence proved

tanx=sinx/cosxcotx=cosx/sinxsecx=1/cosx-5tanx+4sinx-3/sinx+2/cosx-13-5sinx/cosx+2/cosx+4sinx-3/sinx-13-5sinx+2/cosx+4sinx-3/sinx-13

plz give full explanation

The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.so it will be (1/3)²= 1/9

The answer is thatC gets rs.400

2)135 and 225Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45.

1)We know that 504 > 156 so,504=156×3+46156=46×3+1846=18×2+1018=10×1+810=8×1+28=2×4+0Hence, now the remainder has become zero so pur procedure stops and thus we get the HCF = 2

1

Step-by-step explanation:

Consider 1/secA-tanA - 1/cosA

Multiplying by secA + tanA in the numerator and denominator of

first term,

we get secA + tanA/(secA +tanA)(secA - tanA) - 1/cosA

= secA + tanA - secA (Since sec²A - tan²A = 1)

= tanA

Adding and subtracting secA , we get

secA + tanA - secA

= 1/cosA - (secA - tanA)

Now multiplying and dividing (secA - tanA) by (secA + tanA), we

get 1/cosA - (sec²A - tan²A)/(secA + tanA)

= 1/ cosA - 1/secA + tanA

= R.H.S

Hence, Proved.

Hope, it helps !

question not clear. state properly

Inmathematics, and more specifically incalculus,L'Hôpital's ruleorL'Hospital's rule(French:[lopital]) usesderivativesto help evaluatelimitsinvolvingindeterminate forms. Application (or repeated application) of the rule often converts an indeterminate form to an expression that can be evaluated by substitution, allowing easier evaluation of the limit.

related sums please