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( cosec²A = 1 + cot²A)

cosec²A + 1= 1 + cot²A + 1= 2 + (4/3)²= 2 + 16/9= (18+16) / 9= 34/9 = 3 7/9

Let theta = A

LHS = sin A(1+ tan A)+ cos A(1 + cot A)

= sin A + sin^2 A/ cos A + cos A + cos^2 A/ sin A

= sin A + cos A + [sin^3 A + cos^3 A]/ sin A cos A

=[sin^2 A cos A + cos^2 A sin A + sin^3 A + cos^3 A]/sin A cos A

= [sin^2 A cos A +cos^3 A + cos^2 A sin A + sin^3 A]/sin A cos A

= [cos A (sin^2 A + cos^2 A) + sin A (sin^2 A + cos^2 A)]/sin A cos A

= [cos A +sin A]/sin A cos A

= (1/sin A) + (1/cos A)

= cosec A + sec A = RHS.

Hence proved

cosec sqr A - cot sqr A = 1 = (cosec A - cot A)*(cosec A + cot A) ....eq1

cosec A - cot A = p ..... eq 2 given

usin eq 1 we have cosec A + cot A = 1/p ----------- eq3

add eq2 and eq3 we get 2 cosec A = p + 1/ptherefore cosec A = 1/2 (p + 1/p)

Speed= 36km/hr = 10m/secTime= 30 secDistance = Length = Speed * Time = 300 m

¡) +¡¡) -¡¡¡) - answer.

5x² × 4x³ = 20x^ (5)

tan48 = tan(90-42) = cot42 tan23 = tan(90-67) = cot67 tan48*tan23*tan42*tan67 = cot42*cot67*tan42*tan67 = cot42*tan42*tan67*cot67 = 1*1 = 1

a^5 * a^3

= a^8

is the answer

x axisy axisz axis

lsec theta+tan theta(1-sin theta)

1/cos theta + sin theta/cos theta(1-sin theta)

1+sin theta/cos theta(1-sin theta)

1-sin²theta/cos theta

cos²theta/cos theta=cos theta

The degree of the given polynomial is 5.

The degree of a polynomial is the highest degree of its monomials (individual terms) with non-zero coefficients.

Using the following T- functions of sum & difference of angles, we can prove it.

(1) tan 2A = (2tanA) / (1- tan²A)

So, (2) tan 4A = ( 2tan2A) / (1- tan² 2A)

LHS = tanA + 2tan2A + 4/(tan4A)

= tanA + 2tan2A + {4/ (2tan2A)/(1-tan²2A)}….by using (2)nd function

= tanA + 2tan2A + 4(1-tan²2A) / 2tan2A

= tanA + 2tan2A + 2(1-tan²2A) / tan2A

= tanA + {( 2tan² 2A + 2 - 2tan²2A)} / tan2A

= tanA + { ( 2/ tan2A ) }

= tanA + [2 / {2tanA/(1-tan² A)}]…. by using 1st function

= tanA + [ {2( 1-tan²A)} /2tanA ]

= {(2tan² A + 2 - 2tan² A )} / 2tanA

= 2/ (2 tanA)

= 1/tanA

= cot A = RHS

[ Hence Proved]

Soil conservation is a methodology to maintain soil fertility, prevent soil erosion and exhaustion and improve the degraded condition of the soil.

Some commonly used methods of soil conservation are:-

1. Extensive reforestation and afforestation: It is the most effective technique of checking soil erosion .

2. Prevention of faulty practices:

Prevention of over grazing, shifting cultivation, over usage of chemical fertilizers and promoting bio-fertilizers.

3. Changing Agricultural Practices:

- Crop Rotation:This refers to growing of two or more different crops in sequence in a field for maintaining the soil fertility.

- Contour Ploughing:Ploughing parallel to the contours of a hill slope to form a natural barrier for water to flow down the slope.

- Terracing:On steeper slopes, terraces or flat platforms are constructed in steps in a series along the slope. This way water is retained on each terrace which can be used to raise crops. They can reduce surface run-off and soil erosion.

-Trash farmingin which chopped crop residue are spread and ploughed to increase the humus and organic matter content.

4. Check dams: construction of check dams to check the flow of water and reduce it's velocity.

5.Other remedial measures:Water Management , construction of check dams, Wind Break.

thanks for the answer

2+3/4 = 11/4 = 2(3/4)

7/9+1/3 = 7/9+3/9 = 10/9 = 1(1/9)

1-4/7 = (7-4)/7 = 3/7

2(2/3)+1/2 = 8/3+1/2 = (16+3)/6= 19/6 = 3(1/6)

5/8-1/6 = (15-4)/24 = 11/24

2(2/3)+3(1/2) = 8/3+7/2 = (16+21)/6 = 37/6 = 6(1/6)

The sum and difference of 5th powers can be factored as follows:

a5 + b5 = (a + b)(a4 − a3b + a2b2 − ab3 + b4)

a5 − b5 = (a − b)(a4 + a3b + a2b2 + ab3 + b4)

Use protractor and draw 45° then name it as A,B,C from the center cut the line in maths language it's bisect

| 1 w w² || w w² 1 || w² 1 w |

R(1) --> R(1) + R(2) + R(3)

| 1+ w+ w² 1+ w+ w² 1+ w+ w²| | w w² 1 || w² 1 w

| 0 0 0|| w w² 1 || w² 1 w |

So, 0 is answer

European city-state ofMonacois the most densely populated country with a population density of 25,105 people per sq. km (67,612/sq mile). Chinese territory of Macau has the world's 2nd highest population density at 21,151/km².

please send me your geography book name?

this is wrong ANS.

(tan A + cot A=2)=(tan A + cot A=2)^2=tan^2A+2tanAcotA+cot^2=4; tan^2A+cot^2+2=4; tan^2A+cot^2A=2

a)2/√7*√7/√7=2√7/7b)6/√3*√3/√3=6√3/3

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x-2/6x+1 = 1

x-2 = 6x+1

6x-x = -1-2

5x= -3

x = -3/5

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answer is 79079 getted by adding

78879 is ur correct answer

78879 is ur answer . okk

your answer is 78879

78879 is the answer

We have :x+1/x= 2 , we simplify that and get

x^2 + 1 = 2x

x^2 - 2x+ 1 = 0

x^2-x-x + 1 = 0

x(x - 1 ) - 1 (x - 1 ) = 0

(x- 1 ) (x- 1 ) = 0So ,

x =1

Then we substitutex = 1 inx^100+1/x^100, and get

x^100+1/x^100=1^100+1/1^100=1+1=2(Ans)

(1(1/20))^100 (21/20)^100

taking log , => 100log(21/20) = 100×(0.0211) = 2.11

also, log(100) = log(10)² = 2

so, 2.11 > 2

=> (21/20)^100 > 100=> (1(1/20)) > 100

Dividing we get 5920/6 = 986.6667

The expression is (bⁿ)⁴ = (b)ⁿ*⁴ = b⁴ⁿ .

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