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7351.

Cosec31-sec59

Answer» Cosec (90-59)= sec59 [ cosec90- A = sec A]Sec59 - sec 59=0
7352.

Cos (90-theata)× cos theata ÷ tan theata + cos²( 90- theata)= 1

Answer» Cos(90-theata)=sin theataTan theata =sin theata /cos theata Substitute Sin theata.cos that\'s /sin theata /cos theata +sin²theatacos²theata+sin²theata1
7353.

In what ratio is the line segment joining the points P(-2,-3) and Q(3,7)divided by the y-axis?

Answer» Let A (-2, -3) and B (3, 7)P (0, y) and ratio be K : 1Coordinate of P are {tex}\\left( {\\frac{{3k - 2}}{{k + 1}},\\frac{{7k - 3}}{{k + 1}}} \\right){/tex}{tex}\\frac{{3k - 2}}{{k + 1}} = 0{/tex}{tex} \\Rightarrow k = \\frac{2}{3}{/tex}\xa0or 2 : 3
7354.

Express all other trigonometric ratios of tan A

Answer»
7355.

Sin A+sinB

Answer» Sorry by mistake I launched this answer the correct answer is Sin A + Sin B = Sin (A +B) = Sin A Cos B +Sin B + Cos A
Sin A + Sin B = Sin(A+B) = SinA + Cos B
7356.

What is co-prime numbers?

Answer» Those numbers which have no common factor other than 1is called co-prime numbers.
7357.

(1+tan×tan A)+(1+1/tan×tan A)=1/sin×sinA-sin×sin×sin×sinA

Answer» LHS{tex} = \\left( {1 + {{\\tan }^2}A} \\right) + \\left( {1 + \\frac{1}{{{{\\tan }^2}A}}} \\right){/tex}{tex} = \\left( {1 + {{\\tan }^2}A} \\right) + \\frac{{({{\\tan }^2}A + 1)}}{{{{\\tan }^2}A}}{/tex}{tex} = {\\sec ^2}A + \\frac{{{{\\sec }^2}A}}{{{{\\tan }^2}A}}{/tex}\xa0{tex}\\left[ {\\because 1 + {{\\tan }^2}A = {{\\sec }^2}A} \\right]{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A}} + \\frac{{\\frac{1}{{{{\\cos }^2}A}}}}{{\\frac{{{{\\sin }^2}A}}{{{{\\cos }^2}A}}}}{/tex}\xa0{tex}\\left[ \\begin{gathered} \\because {\\sec ^2}A = \\frac{1}{{{{\\cos }^2}A}} \\hfill \\\\ {\\tan ^2}A = \\frac{{{{\\sin }^2}A}}{{{{\\cos }^2}A}} \\hfill \\\\ \\end{gathered} \\right]{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A}} + \\frac{1}{{{{\\cos }^2}A}} \\times \\frac{{{{\\cos }^2}A}}{{{{\\sin }^2}A}}{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A}} + \\frac{1}{{{{\\sin }^2}A}}{/tex}{tex} = \\frac{{{{\\sin }^2}A + {{\\cos }^2}A}}{{{{\\cos }^2}A{{\\sin }^2}A}}{/tex}{tex} = \\frac{1}{{{{\\cos }^2}A{{\\sin }^2}A}}{/tex}\xa0{tex}\\left[ {\\because {{\\sin }^2}A + {{\\cos }^2}A = 1} \\right]{/tex}{tex} = \\frac{1}{{(1 - {{\\sin }^2}A){{\\sin }^2}A}}{/tex}\xa0{tex}\\left[ {\\because {{\\cos }^2}A = 1 - {{\\sin }^2}A} \\right]{/tex}{tex} = \\frac{1}{{{{\\sin }^2}A - {{\\sin }^4}A}}{/tex}= RHSHence proved.
7358.

Can you please send me the formulas of chapter trigonometry

Answer» You can check revision notes for formulae :\xa0https://mycbseguide.com/cbse-revision-notes.html
7359.

If tanA=√2-1 show that sinAcosA=√2\\4

Answer»
7360.

Prove that /2+/3 is an irrational number

Answer» Let us assume that\xa0{tex}\\sqrt 2 + \\sqrt 3{/tex}\xa0is a rational numberLet {tex}\\sqrt2+\\sqrt3=\\frac{\\mathrm a}{\\mathrm b}{/tex} Where a and b are co-prime positive integersOn squaring both sides, we get{tex}(\\sqrt2+\\sqrt3)^2=\\frac{\\mathrm a^2}{\\mathrm b^2}{/tex}{tex}2+3+2\\sqrt6=\\frac{\\mathrm a^2}{\\mathrm b^2}{/tex}{tex}5 + 2\\sqrt 6 =\\frac{a^2}{b^2}{/tex}\xa0{tex}2\\sqrt6=\\frac{\\mathrm a^2}{\\mathrm b^2}-5{/tex}{tex}2\\sqrt6=\\frac{\\mathrm a^2-5\\mathrm b^2}{\\mathrm b^2}{/tex}{tex}\\sqrt6=\\frac{\\mathrm a^2-5\\mathrm b^2}{2\\mathrm b^2}{/tex}Now\xa0{tex}\\frac{\\mathrm a^2-5\\mathrm b^2}{2\\mathrm b^2}{/tex}\xa0is a rational number.This shows that\xa0{tex}\\sqrt 6{/tex}\xa0is a rational number.But this contradicts the fact that\xa0{tex}\\sqrt 6{/tex}\xa0is an irrational number.This contradiction has raised because we assume that\xa0{tex}\\left( {\\sqrt 2 + \\sqrt 3 } \\right){/tex}\xa0is a rational number.Hence, our assumption is wrong and\xa0{tex}\\left( {\\sqrt 2 + \\sqrt 3 } \\right){/tex}\xa0is an irrational number.
7361.

Find the value of x when the value of 2x is 98.

Answer» Yes it is 49
Ya ya it\'s 49.
Ya you are right
if 2x=98then, x= 98/2i.e x=49
49
49
7362.

under root 5 is irrational number

Answer» Yes it is an irrational no. In exams u have to prove this too
You have to prove it also in board exams .
Koi shak
7363.

√2+√2+√2+√2=?

Answer» Ya the value of root 2 is 1.41 so there was 4 root2 so multiply 4 by 1.41 then answer is 5.656. Ok!
√2+√2+√2+√2= 4√2=4×1.41(√2= 1.414)=5.656This is the answer
7364.

Please easy way to learning that question

Answer»
7365.

Middle term spliting

Answer»
7366.

Plz! Tell me easy way to learn trigonometry formulas

Answer» sin 0= 0/4 rootsin 30= 1/4rootsin 45= 2/4rootsin 60= 3/4rootsin 90= 4/4root
Thre was a another way that the sin,theta equal to perpendicular upon hypotenuse..In all the fomulas you know may be so the trick is PBP HHB ,first there is p upon h means perpendicular upon hypotenuse.Second b upon h base upon hypotenuse.Third p upon b perpendicular upon base .In fourth you have to take it inversely h upon p hypotenuse upon perpendicular. like you can do h upon b and p upon b.oo i forgot tell that from upper side you take that sin cos tan pbp k upar likho. And then hhb k niche likho cosec sec cot.to phir dekho upar likha sin to sin ka ho gya p upon h phor cos uska b upon h phir likha p upon b to uska tan phir niche aao to h upon p uske niche likha tha cosec phir h upon b sec phir b upon p to uska cot ok its too much long ☺☺☺☺but read it hole.
Make some rhymes on it Or do like this that see t comes after s then u can easily remember the formula of sec and tan queta
7367.

Which angle make pie

Answer»
7368.

A number

Answer» 11
1
7369.

How many three digit numbers divisible by 3?

Answer» Infinite
7370.

What is a frustom of a cone

Answer» Not from upper part but down part is the area where it is cut from and it somewhat looks like a bucket
If cut the part of a cone from upper side then it becomes the frustum.
7371.

Find an a.p for these 3,5,7,9

Answer»
7372.

If the sum of first nth termof an A.P isa{n}^{2}+bn find its common difference\u200b

Answer»
7373.

Evaluate 2% Of 2%

Answer» 0.4
4
7374.

cos theta - sin theta + 1 / cos theta + sin theta -1

Answer»
7375.

What is relative prime

Answer» Two numbers are "relatively prime" when they have no common factors other than 1.In other words you cannot evenly divide both by some common value.Examples: 7 and 20 are relatively prime (no common factor)6 and 20 are not relatively prime because you can evenly divide both by 2 (2 is a common factor).
7376.

Find hcf of 3³×5³

Answer» 1
7377.

2 Ka power 100

Answer» {tex}{\\left( {{2^{10}}} \\right)^{10}} = {\\left( {1024} \\right)^{10}}{/tex}
7378.

11+11=412+12=913+13=? Give answer

Answer» Rahul u are wrong dear.
15
7379.

5:7=

Answer»
7380.

One hectare=?

Answer»
7381.

_tanA_____ =_______tanA-1______1+cotA 2-consec square A

Answer»
7382.

ए प्लस बी का होल क्यूब

Answer» a^3 +b^3 +3ab(a+b)
?on Google
7383.

Is 5 is a rational no.

Answer» Yes
Yes 5/1
7384.

Kya board ka paper NCERT say ayega

Answer»
7385.

Sir class ten final examination ma mcq aiga kya

Answer»
7386.

Hello sir

Answer»
7387.

How is quaratic equation formed if we have given sum of zeros and products of zeros?

Answer» k{x2-(alpha +beta )x +alpha beta}
7388.

Cos45/sec30+cosec30

Answer»
7389.

Is the final exam is hard?

Answer»
7390.

Application of LCM and HCF

Answer»
7391.

The HCF of two numbers in 27 and LCM is 162.If one of the number is 81,Find the other

Answer» 54
7392.

If a and b are

Answer» To b aur c to jaroor honge d aur e bh....
7393.

5x+3y=02x+4y=5

Answer»
7394.

1/v + 1/u =1/f is the formula of

Answer» Here is the right answer
Thats Mirror Formula.
7395.

If alpha and beta are the zeroes of polynomial x2-5x+k and a-p =1 find the value of k

Answer» 6
7396.

If the sum of first \'m\' terms of an AP is n,then show that the sum of its first (m+n)=-(m+n)

Answer» Let a be the first term and d be the common difference of the given AP. Then,Sm = n\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{m}{2}{/tex}[2a + (m-1)d] = n{tex}\\Rightarrow{/tex}\xa02am + m(m- 1)d - 2n\xa0...... (i)And, Sn\xa0= m\xa0{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a + (n - 1)d] = m{tex}\\Rightarrow{/tex}\xa02an + n(n - 1)d = 2m ...... (ii)On subtracting (ii) from (i), we get2a(m-n) + [(m2 - n2) - (m - n)]d = 2(n - m){tex}\\Rightarrow{/tex}\xa0(m - n)[2a + (m + n - 1)d] = 2(n - m){tex}\\Rightarrow{/tex}\xa02a + (m + n- 1)d = -2 ..... (iii)Sum of the first (m + n) terms of the given AP=\xa0{tex}\\frac{{(m + n)}}{2}{/tex}{tex}\\cdot{/tex}{2a + (m + n - 1)d}{tex}= \\frac { ( m + n ) } { 2 } \\cdot ( - 2 ) = - ( m + n ){/tex}\xa0[using (iii)].Hence, the sum of first (m + n) terms of the given AP is -(m + n).
7397.

If m times the nth term of an AP is same as n times mth term , find the (m+n)th term

Answer» Let a be the first term and d be the common difference of the given AP. Then,Sm = n\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{m}{2}{/tex}[2a + (m-1)d] = n{tex}\\Rightarrow{/tex}\xa02am + m(m- 1)d - 2n\xa0...... (i)And, Sn\xa0= m\xa0{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a + (n - 1)d] = m{tex}\\Rightarrow{/tex}\xa02an + n(n - 1)d = 2m ...... (ii)On subtracting (ii) from (i), we get2a(m-n) + [(m2 - n2) - (m - n)]d = 2(n - m){tex}\\Rightarrow{/tex}\xa0(m - n)[2a + (m + n - 1)d] = 2(n - m){tex}\\Rightarrow{/tex}\xa02a + (m + n- 1)d = -2 ..... (iii)Sum of the first (m + n) terms of the given AP=\xa0{tex}\\frac{{(m + n)}}{2}{/tex}{tex}\\cdot{/tex}{2a + (m + n - 1)d}{tex}= \\frac { ( m + n ) } { 2 } \\cdot ( - 2 ) = - ( m + n ){/tex}\xa0[using (iii)].Hence, the sum of first (m + n) terms of the given AP is -(m + n).
7398.

prove that 2root3÷5 is irrational

Answer» Let2√3/5 be rational no.then 2√3/5 is in form of a/bWhere, a & b are co prime2√3/5=a/b√3=5a/2bHere,5a/2b is rationalBut,√3is irrationalThis is contradictioHence,2√3/5 is irrational
7399.

Can two number have 18as their HCF and 380 as their LCM ? Give reason .

Answer» NO because hcf doesnot completely divides lcm
7400.

In triangle POR XY equal QR if XY=1/3QR than find PR:PX.

Answer»