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2101.

sinx**3 - cosx**3 is equals to

Answer»
2102.

Maths ch 3 extras questions

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2103.

Unit 1 miscellanous exercise

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2104.

why sinx+sin5x=sin6x

Answer» Sin (x+5x)=sin6x
Sin (x+5x)Sin6x
Sina +sinb indentity
2105.

Find the value of cos 765°

Answer» 1/√2
-1/root 2
Cos (360×2+45)Cos (45)Cos (1/root2
1/√2
2106.

Venn diagram

Answer»
2107.

Prove tan3x tan2x tanx= tan3x - tan2x - tanx

Answer» Eg. Dekh le bc
2108.

How can solve mode question f(x)=-|x|

Answer» F (x)= -x
Agar mod ke under koi sign rehe toh woh plus ho jata hai ,,agar -( minus) rahega toh plus agar plus toh plus hi hoga .Agar bahar koi sign rahe toh wahi sign hoga
-x hoga
2109.

Cot2π/6+cosec5π/6+3ttan2π/6

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2110.

Chapter 8 elements solutions in

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2111.

Underoot 2x-1 upon x-1 less than zero

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2112.

f(x)=√x-3 find domain and range

Answer» x-3≥0x≥3Domain = [3,infinity)For range let y=√x-3ysquare =x-3x=y square+3 So Range=[0,infinity)
2113.

kya koi pratibha school se hai

Answer» ?????are zubair tum..kaise ho
2114.

Please explain the questions of 3.3 question number 23 and 24 by a video

Answer»
2115.

Prove this eq cosx(pi/2-x)=sinx

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2116.

(2n+7)>(n+3)² prove by mathematical induction

Answer» Let P (n) =(2n + 7) < (n + 3)2For n = 1{tex}P(1) = (2 \\times 1 + 7) < {(1 + 3)^2} \\Rightarrow 9 < 16{/tex}{tex}\\therefore {/tex}\xa0P ( 1) is trueLet P(n) be true for n = k{tex}\\therefore P(k) = (2k + 7) < {(k + 3)^2}{/tex}\xa0....(1)For n = k + 1\xa0P (k + 1) = 2 (k + 1) + 7 < (k + 1 + 3)2{tex} \\Rightarrow {/tex}\xa02(k + 1) + 7 < (k + 4)2From (1)2k + 7 < (k + 3)2Adding 2 on both sides2k + 7 + 2 < (k + 3)2 + 2\xa0{tex} \\Rightarrow {/tex}\xa02(k + 1) + 7 < k2 + 9 + 6k + 2{tex} \\Rightarrow {/tex}\xa02(k + 1) + 7 < k2 +6k + 11 < k2 + 8k + 162(k + 1) + 7 <(k + 4)2{tex}\\therefore {/tex}\xa0P (k + 1) is trueThus P(k) is true {tex} \\Rightarrow {/tex}\xa0P(k + 1) is trueHence by principle of mathematical induction, P (n) is true for all {tex}n \\in N{/tex}.
2117.

Exercise 9.2 question number 9

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2118.

The angle between the minute and hour hands of a clock at 8:30is

Answer» Sorry answer is wrong In 30 minutes =180° So 255-180=75°
12hours = 360° 1 hour = 360÷128:30 hours = 8+1÷2×360÷12 = 255°-----------(i) Now 60minute = 360°1minute =360÷60=1÷630 minute = 30×1÷6=5°----------------------(ii) Substract equation ii by i255°-5°=250°
May 60 degree or 300 degree
270
75°
75 degree
2119.

Find the range of f(x)= 1/√9-x^2

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2120.

Solve the equation-|z| = z+1+2i

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2121.

What is the definition of set ??

Answer» Well defined collections of distinct objects is called set
Well defined collection of objects
A set is a collection of well - defined objects or elements
2122.

Tan3x tan2x tanx=tan3x-tan2x-tanx

Answer» Abhijeet Sharma...can u pls explain ur steps in the ans u submitted...
>>Tan3x=tan(2x+x)>>Tan3x=tan2x+tanx/1-tan2x•tanx>>Tan3x(1-tan2x.tanx)=tan2x +tanx>>Tan3x-Tan3xTan2xtanx=tan2x+tanx>>Tan3xtan2xtanx=tan3x-tan2x-tanx
2123.

Prove that :a+b/c=cos(a-b/2)/sinc/2

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2124.

Find square root of ( -i) of complex Number

Answer» Let {tex}x + yi = \\sqrt { - i} {/tex}Squaring both sides, we get(x + yi)2 = -ix2 - y2 + 2xyi = -iEquating the real and imaginary partsx2 - y2 = 0 . . . (i)and 2xy = -1{tex}\\therefore xy = - \\frac{1}{2}{/tex}Now using the identity(x2 + y2)2 = (x2 - y2)2 + 4x2y2{tex} = {(0)^2} + 4{\\left( { - \\frac{1}{2}} \\right)^2}{/tex}= 1{tex}\\therefore {/tex}\xa0x2 + y2 = 1 . . . (ii) [Neglecting (-) sign as x2 + y2 > 0]Solving (i) and (ii), we get{tex}{x^2} = \\frac{1}{2}{/tex}\xa0and {tex}{y^2} = \\frac{1}{2}{/tex}{tex}\\therefore x = \\pm \\frac{1}{{\\sqrt 2 }}{/tex}\xa0and {tex}y = \\pm \\frac{1}{{\\sqrt 2 }}{/tex}Since the sign of xy is (-) then if{tex}x = \\frac{1}{{\\sqrt 2 }}{/tex},\xa0{tex}y = - \\frac{1}{{\\sqrt 2 }}{/tex}If {tex}x = - \\frac{1}{{\\sqrt 2 }}\\;y = \\frac{1}{{\\sqrt 2 }}{/tex}{tex}\\style{font-family:Tahoma}{\\style{font-size:8px}{\\therefore\\sqrt{-i}=\\pm\\left(\\frac1{\\sqrt2}-\\frac1{\\sqrt2}i\\right)}}{/tex}
2125.

COS20 COS40 COS80-1/8

Answer» So 1/8 -1/8 is 0
The value of cos20 cos40 cos80 is 1/8
0
2126.

If (x/3+1,y-2/3)=(5/3,1/3),find the values of x and y.

Answer» X=2 and y=1
X=2 and y=-1/2.
2127.

Cos x =3/5,x lies in second quadrant

Answer» Cos x= -3/5 hoga because cos thetha is negative in 2nd quadrant.
Instead of cosx it should be sin x
2128.

3f(x)+2f(1/x)=x*x

Answer»
2129.

(A+b)( a-b)

Answer» a2- b2
a^2-b^2
(a + b)( a - b) = a2 - b2
A square-B square
A square ×B square
2130.

2sinx +2 cosx =1

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2131.

Prove that 0!=1Where (!)=factorial

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2132.

Make modulus function graph for -4,4

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2133.

since (B+C) ÷2=(b-c) ÷a×cos(A) ÷2

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2134.

Let Sn denotes the sum of n terms of an A.P. if S2n =3Sn , the ratio S3n / sn=

Answer» Let 1st term be a and common difference be d.Given, S2n\xa0= 3sn{tex}\\therefore{/tex}\xa0{tex}\\frac { 2 n } { 2 }{/tex}\xa0[2a + (2n - 1) d] = 3 {{tex}\\frac { n } { 2 }{/tex}\xa0[2a + (n - 1) d]}{tex}\\Rightarrow{/tex}\xa04a + (4n - 2) d = 6a + (3n - 3)d{tex}\\Rightarrow{/tex}\xa02a = (n + 1)d\xa0...(i)Now,\xa0{tex}\\frac { S _ { 3 n } } { S _ { n } } = \\frac { \\frac { 3 n } { 2 } [ 2 a + ( 3 n - 1 ) d ] } { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] }{/tex}=\xa0{tex}\\frac { 3 \\{ ( n + 1 ) d + ( 3 n - 1 ) d \\} } { ( n + 1 ) d + ( n - 1 ) d }{/tex}\xa0[from Eq. (i)]=\xa0{tex}\\frac { 3 ( 4 n d ) } { 2 n d }{/tex} = 6:1
2135.

The sum of numbers from 250 to 1000 which is divisible by 3 is

Answer» 45°
2136.

Write the value of tan 3π/12

Answer» Tan 3π/12=3π/12×180/π=45°tan45°=.
1
Answer is 1
Answer is 1
It is 1, if you will write it as tan(π/4) then ot will look easy.... Hope it may help you...
2137.

Find tan5x in term of tan

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2138.

Find the rank of the word "VARTIKA" and "ANNU"

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2139.

Cos(60+a) sin(60-a)=1/4(root3-2sin a)

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2140.

Trigononetric formulas used in chapter 3 of ncert

Answer» Chapter is Trigonometric Functions
Dekh iss app me maths me ja aur lesson trigo choose kar all formula diya hua hai . Ok
Can you please tell me the full name of the chapter so that I can help you because I am not having the book with me
It may be formula of compound angle
2141.

In how many ways can an examinee answer a set of 10 true false type questions

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2142.

6π÷0 =

Answer» Friend it actually it is not defined but we consider it as infinity. Hope it may help you...
Zero
2143.

Prove that 1+cossquare 2theta = 2 ( cos power 4theta +sinpower 4 theta)

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2144.

The sum of first 7 terms of an AP is 10 and that of next 7 terms is 17. Find the progression.

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2145.

1+(1+3)+(1+3+5)+...+ n terms=n(n+1)(2n+1)/6

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2146.

a+(a+d)+(a+2d)+...+ a+(n-1)d =n/2 2a+(n-1)d

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2147.

Exercise - 4.1 , Q - 1

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2148.

(1+w)^3-(1-w^2)^3

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2149.

1w((1+w)

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2150.

Class 11 maths chapter 1 mescellious exercise question 8 , 10 ,11,12 please answer

Answer»