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As we know A + B + C = 180So, B + C = 180 - AOr, (B/2) + (C/2) = 90 - (A/2) ....(1)

Now think in triangle BOC,we have three angles, OBC, BOC and OCBOBC = (B/2)OCB = (C/2)Let BOC = X

So, X + (B/2) + (C/2) = 180From equation (1)

X + 90 - (A/2) = 180X = 90 + A/2 (Proved !!!)

Sin34° + Cos64° - Cos4°= Sin34° + Cos(34 + 30)° - Cos(34 - 30)°= Sin34° + (-2Sin34°Sin30°)= Sin34° - 2 × Sin34° × 1/2= Sin34° - Sin34°= 0 (Proved)

Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(x - π/3) = {1 + cos(2x - 2π/3)}/2

ii) Hence, left side of the given one is:

= {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2

= (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)]

= (3/2) + (1/2)[cos(2x) + 2cos(2x)*cos(2π/3)][Since cos(A+B) + cos(A-B) = 2cosA*cosB]

= (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2]

= 3/2 = Right side HENCE PROVED

Given:A = {3, 6, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}i A−BÂ = {3, 6, 15, 18, 21}ii A−C = {3, 15, 18, 21}iii A−D = {3, 6, 12, 18, 21}iv B−A = {4, 8, 16, 20}v C−A = {2, 4, 8, 10, 14, 16}vi D−A = {5, 10, 20}vii B−C = {20}viii B−D = {4, 8, 12, 16}

thanks sir

120=2*2*2*3*5504=2*2*2*3*3*7882=2*3*3*7*7

so LCM=2*2*2*3*3*5*7*7=17640

first second ke rule se lagaiye sir aa jayega pahle ko second aur dusre ko first

make substitution of t = arctan((x)^3))herearctan refers to tan inverse

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CosA/sinA - sinA/cosA=(2cos2A-1)/sinAcosAlhs:cos^2A-sin^2A/sinAcosA cos^2A-1+1-sin^2A/sinAcosAcos^2A-1+cos^A/sinAcosA 2cos^2a-1/sinAcosA=rhs

Pleaseeeeeeeee33 like

answer is (B) 2.5 cmplease like my answer thanks

answer is (B) 2.5 cm please like my answer

option (b) 2•5 is the right answer

ANSWER IS 2. 5cm (option b)

no. of zeroes are 3

no. of zeroes of p(x) : 3

Thanku😊

As we know that tan(90- theta) = cot theetaso tan(90-65} = cot 25°cot 25°/cot 25°so answer= 1please let me know if you need any helplike it

Idk who you would juqqkiqkiq

y= x^2 + 1/y

y^2 = x^2 y +1

differentiate both side

2y y' = 2xy + x^2 y'

2yy' - x^2 y' = 2xy

y'( 2y - x^2) = 2xy

y'= 2xy/(2y - x^2)

xcosθ = 1ysinθ = 1

ysinθ/xcosθ = 1ytanθ/x = 1tanθ = x/y

c) is the correct answer

you are great 😊😊😊

Please accepted as best please

the gas which turns lime milky is carbon dioxide

Thanks for helping

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