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James Mill published a three-volume work, AHistoryofBritish Indiain 1817. He was a Scottish economist and political philosopher. In his work, hedivided the Indian historyinto three periods of Hindu, Muslim andBritish.Historianshavedivided the Indian historyinto- ancient, medieval, modern, colonial.

LHS = tan4x = tan(2x + 2x) use, the formula, tan(A + B) = (tanA+tanB)/(1-tanA.tanB)

= (tan2x + tan2x)/(1-tan2x.tan2x) =2tan2x/(1-tan²2x)

again, use the formula, tan2A = 2tanA/(1-tan²A)

= 2{2tanx/(1-tan²x)}/[1-{2tanx/(1-tan²x)}²]=4tanx.(1-tan²x)²/(1-tan²x)(1+tan⁴x-2tan²x-4tan²x)=4tanx.(1-tan²x)/(1-6tan²x+tan⁴x) = RHS

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Give me solution for following:

For 🔺 ABC a=18, b=24, c=30 Find the value of cos A

From 3x = 2x + xApplying tan on both sides..tan3x = tan (2x+x)tan3x = (tan2x+tanx)/1-tan2x tanx(1-tan2x tanx)tan3x = tan2x+tanxtan3x - tanx tan2x tan3x = tan2x+tanxtanx tan2x tan3x = tan3x-tan2x-tanx

Hence it is proved..

thanks

secx=2

⇒ cos x = 1/2

cos x = cosπ/3

Using formula for general soluton of cos x = cos y i.e x = 2nπy

We have x = 2nπ-π/ 3

The solutions that lie in the interval 0 to 2π, both included are principal solution

So they are , in this case ,

π/3, 2π-π/ 3 = 5π/ 3

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.

Multiplying by cos(x/2), both numerator and denominator,{1 - tan(x/2)}/{1 + tan(x/2)} = {cos(x/2) - sin(x/2)}/{cos(x/2) + sin(x/2)}

2) Then rationalizing, by multiplying with {cos(x/2) - sin(x/2)}{1 - tan(x/2)}/{1 + tan(x/2)} = {cos(x/2) - sin(x/2}²/{cos²(x/2) - sin²(x/2)}= {cos²(x/2) + sin²(x/2) - 2sin(x/2)*cos(x/2)}/{cos(x)} [Expanding the numeration using (a-b)² identity and by double angle identity denominator = cos(x)]This further simplifies to: {1 - sin(x)}/cos(x) [Since by identity, cos²A + sin²A = 1 and by multiple angle identity, sin(2A) = 2sin(A)cos(A)]

3) Thus the question simplifies to: {1 - sin(x)}²/[{cos(x)}*(π - 2x)³]

4) Let π - 2x = t; So, x --> π/2, t --> 0.Also x = π/2 - t/2==> cos(x) = cos(π/2 - t/2) = sin(t/2) = 2sin(t/4)cos(t/4)

As well, 1 - sin(x) = 1 - sin(π/2 - t/2) = 1 - cos(t/2) = 2sin²(t/4) [By multiple angle identity, {1 - cos(2x) = 2sin²x]==> {1 - sin(x)}² = 4sin⁴(t/4)

5) As of the above, the question transforms to:Lim (t --> 0) [{4sin⁴(t/4)}/{2sin(t/4)cos(t/4)(t³)}]

This simplifies to: Lim (t --> 0) [2sin³(t/4)/{(t³)*cos(t/4)}]

= 2*Lim (t --> 0) sin³(t/4)/t³ [As t --> 0, cos(t/4) = cos(0) = 1]

= (2/64)*Lim (t --> 0) [{sin(t/4)/(t/4)}³]

= (2/64)*1 = 1/32

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1080 = (3000×12×T)/100 T = 108/36 = 3 वर्ष

Thankyou

thank u very much but answer is ₹2721.90

No of flower bed=5each measure=1marea=a² =1m²total area of flower bed=5m²area of land=l×b=5×4=20m²remaining part=20-5=15m²

Given:Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)

Diagonal of a square (AC) =√2×side of a square.

Diagonal of a square (AC) =√2 × 56 = 56√2 m.OA= OB = 1/2AC = ½(56√2)= 28√2 m.[Diagonals of a square bisect each other]

Let OA = OB = r m (ràdius of sector)

Area of sector OAB = (90°/360°) πr²

Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m²

Area of sector OAB = 1232 m²

Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²

Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m²

Area of flower bed AB = 448 m²

Similarly, area of the other flower bed CD = 448 m²

Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m²

Hence, the sum of the areas of the lawns and the flower beds are 4032 m².

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Given:Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)

Diagonal of a square (AC) =√2×side of a square.

Diagonal of a square (AC) =√2 × 56 = 56√2 m.OA= OB = 1/2AC = ½(56√2)= 28√2 m.[Diagonals of a square bisect each other]

Let OA = OB = r m (ràdius of sector)

Area of sector OAB = (90°/360°) πr²

Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m²

Area of sector OAB = 1232 m²

Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²

Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m²Area of flower bed AB = 448 m²

Similarly, area of the other flower bed CD = 448 m²

Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m²

Hence, the sum of the areas of the lawns and the flower beds are 4032 m².

(c)a-b=b-a is incorrect answer

(c) a-b=b-a is incorrect option

c. a-b=b-a is incorrect answer ..........let a=4 b=2a-b=4-2=2b-a=2-4=-2Both are unequal

A+B=B+A. ..

c , incorrect because a-b=-b+a

c option is not correct otherwise all are shai h

b-a is incorrect answer

the (c) option ie :- a-b=b-a is correct

this (c) option ie :-a-b=b-a this correct Ans

(c) a-b=b - a, wrong❌ he

option c is correct answer

a-b=b-a is incorrect

atan+bsec=c

bsec=c-atan

squaring bothsides

b^2sec^2=(c-atan)^2

b^2(1+tan^2)=c^2+a^2tan^2-2actan

(b^2-a^2)tan^2+2actan+b^2-c^2=0

since alpha and beta are the roots of the eq so

tan(alpha)+tan(beta)=-2ac/(b^2-a^2)

tan(alpha).tan(beta)=(b^2-c^2)/(b^2-a^2)

tan(alpha+beta)={tan(alpha)+tan(beta)}/{1-tan(alpha).tan(beta)}

=[-2ac/(b^2-a^2)]/[1-(b^2-c^2)/(b^2-a^2)]

=-2ac/(c^2-a^2)

=2ac/(a^2-c^2) Ans.

it's the best answer.If you I like it.you can loved it..,.........,..........

sorry last line main disturb hoke bhul Kiya hoon sahi kar lo you can understand that ok......

-20 hoga ...k maine galat se 20likha hun ...k

hit like if you find it useful

1.Draw a line segment AB=6cm2.Draw angle A=603.Draw an arc from B with radius 7cm which cuts the previous line at C4.Draw bisector PQ for BC5.Draw bisector YZ for AC6.From O with radius OC draw a circle

y=x+52.........1adjecent angle72=x+52x=20y=72

here l is parallel to mso y=72and x+52=72 so x=20