Explore topic-wise InterviewSolutions in Current Affairs.

Thanks

Please like the solution 👍 ✔️

We have to prove that Cos2A/1+sin2A=tan(π/4-A)

Solution:-As shown in attachment...

Formula used :Cos 2x = cos²x- sin²x Sin²x +cos² x =1Sin 2x = 2sin x cos x And (1-tan x)/(1+tan x) = tan (π/4 -x)

A = {a,b,c,d}

P(A) = { 0, {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d} {b,c}, {b,d}, {c,d}, {a, b,c}, {b,c,d}, {a,b,d}, {a,c,d}, {a,b,c,d} }

n(P(A)) = 2^4 = 16

2√3*2√3-9*2=12-18=-6

tq....

tq....

The result would be non terminating and repeating because denominator contains the factor of 3^2 (from 6^2) which does not divides the numerator hence does not terminates and repeates.

0 because there are only 52 weeks in a year

Solution:1 year = 365 daysA leap year has 366 daysA year has 52 weeks. Hence there will be 52 Mondays for sure.52 weeks = 52 x 7 = 364 days366 – 364 =2 daysIn a leap year there will be 52 Mondays and 2 days will be left.These 2 days can be:Sunday, MondayMonday, TuesdayTuesday, WednesdayWednesday, ThursdayThursday, FridayFriday, SaturdaySaturday, SundayOf these total 7 outcomes, the favourable outcomes are 2.Hence the probability of getting 53 Mondays in a leap year = 2/7.

(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cos A

multiply LHS by cosA /cosA to get(sinA+1-cosA) / (sinA-1+cosA)

multiply again by cosA/cosA to get(sinA.cosA+cosA-cos^2A) / cosA(sinA-1+cosA)

= ( cosA(1+sinA) - (1-sin^2A) ) / cosA(sinA-1+cosA)= ( cosA(1+sinA) - (1+sinA)(1-sinA) ) / cosA(sinA-1+cosA)= ( (1+sinA)(cosA-1+sinA) ) / cosA(sinA-1+cosA)= (1+sinA)/cosA

p(a)=x^2+x-6 and f(a)=x^2+3x-15; p(a)=f(a); p(a)=x^2+x-6 =x^2+3x-15=f(x);; x-6=3x-15 3x-x=15-6; 2 x=9; x=9/2

The image you have submitted is not cropped properly. Kindly upload complete question.

Ans :- Acoefficientis usually a constant quantity, but thedifferential coefficient of f is a constantfunctiononly if f is a linearfunction. When f is not linear, its differential coefficientis afunction, call it f′, derived by thedifferentiation of f, hence, the modern term, derivative.

fff

the value of the x according to the question is option ( b) 3

right anawer is; (b) 3

x=2+root(3)so 1/x=2-rot(3)so x+1/x=4so (x+1/x)^2=x^2+1/x^2+2=16so x^2+1/x^2=14so (x+1/x)^3=64so x^3+3x+3/x+1/x^3=64so x^3+1/x^3=64-3(x+1/x)=64-3(4)=64-12=52

7root(3)/root(10)+root(3) - 2root(3)/root(6) + root(5) - 3root(2)/root(15) + 3root(2)

= [7root(3)*(root(10)-root(3))/(10 - 3)] - [2root(3)*(root(6)-root(5))]/(6 - 5)] - [3root(2)*(root(15) - 3root(2))/(15 - 18)]

= (root(30) - 3) + (6root(2)-2root(15) - (-root(30) + 18)

= 2root(30) - 2root(15) + 6root(2) - 21

If you find this solution helpful, Please give it a 👍

thank you👍

ty

Radius of cylinderical pillars = r = 28cm =0.28mheight ,h=4mcurved surface area of a cylinder =2πrhcurved surface area of a pillar= 2×22\6×0.28×4=7.04metresq.Curved surface area of 24 such pillars = 7.04×24=168.96metresq.cost of cementing an area of 1metresq. =Rs8so,cost of painting 1689.6metresq.=168.96×8=Rs1,351.68

From which book have you taken this question? Please tell us so that we can provide you faster answer.

derivative( differential) of a function of one variable by the increment of the independent variable.

nice

agar base same ho to division me power subtract ho jata hai (-7)^32/(-7)^3 = (-7)^32-3 = (-7)^29

(-7)^32÷(-7)^3=-7^32-3=-7^29

7^29 is the right answer

left hand side L.H.S3^2x-7so, we let x as -9 ok2-9-7=03^0=1 right hand side R.H.S1 left hand side equals right hand side so X = - 9

what is neutralisation ?

Area of pathway is 88m^2Let R be radius of outer circle =8mLet radius of inner circle be rArea of pathway be 88π(R^2-r^2)=88(8)^2-r^2=28r^2=64-28r^2=36r=6m

SecA+tanA=p ----------------------------(1)∵, sec²A-tan²A=1or, (secA+tanA)(secA-tanA)=1or, secA-tanA=1/p -----------------------(2)Subtracting (2) from (1) we get,2tanA=p-1/por, tanA=(p²-1)/2p∴, cotA=2p/(p²-1)Now, cosec²A-cot²A=1or, cosec²A=1+cot²Aor, cosec²A=1+{2p/(p²-1)}²or, cosec²A=1+4p²/(p²-1)²or, cosec²A=(p⁴-2p²+1+4p²)/(p²-1)²or, cosec²A=(p⁴+2p²+1)/(p²-1)²or, cosec²A=(p²+1)²/(p²-1)²or, cosecA=(p²+1)/(p²-1)

4(7y+33)=3(9y+60)28y+132=27y+180y=48

87/27= as 27*3= 81so 3 and remainder= 6

98/32= 32*3= 96so quotient = 3remainder = 2

Here 60°+5x+5x=180°60°+10x=180°10x=120°x=12°

32/9 = 3.55..34/9 = 3.77...so option (d) 11/3 = 3.666.. is between them

sin = cossin/cos = 1tan = 1so = 45°therefore 2tan²(45°)+ sin²(45°)2×1² + (1/√2)²2 + 1/25/2 option (B) is correct