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evaluate {tex}evaluate sigma superscript 13 subscript n=1 i^n + i^n+1) where n belongs to N{/tex}

Posted by Bahirithi Karempudi (Apr 22, 2017 11:23 a.m.) (Question ID: 4991)

Write the following sets in roster form

1) B = {x : x is a positive factor of a prime number}

2) C={x : x2=x where x belongs to Real numbers}

Answer it quickl

Fast.pls its very urgent..

Posted by Anshika Solanki (Apr 20, 2017 9:11 p.m.) (Question ID: 4975)

Write the following sets in roster form

1) B = {x : x is a positive factor of a prime number}

2) C={x : x2=x where x belongs to Real numbers}

Answer it quickly

Posted by Anshika Solanki (Apr 19, 2017 10:36 p.m.) (Question ID: 4956)

   If AunionB=AintersectionB then proved A=B

Posted by Rupam Ghosh (Apr 11, 2017 4:12 p.m.) (Question ID: 4833)

F(x) = x÷1+x²

Posted by rishab mittal (Mar 30, 2017 8:58 p.m.) (Question ID: 4620)

  • Answers:
  • X÷1+x*

    2x^

     

    Answered by Abhishek Bhardwaj (Mar 31, 2017 7:41 a.m.)
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if f is a function satisfying f(x+y)=f(x) f(y) for all x,y belongs to n such that f(1)=3 and sigma x =1 to n f(x)=120 then find n.

Posted by Koratamaddi Vamsi (Mar 30, 2017 7:03 p.m.) (Question ID: 4617)

  • Answers:
  • Ans. Given: f(x + y) = f(x)*f (y) for all x, y {tex}\in {/tex} N ......(1)

    {tex}\displaystyle\sum_{i=1}^{n} f(x) = 120 \space \space ............(2){/tex}

    f(1) = 3

    Taking x = y = 1 in (1), we obtain

    f(1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

    Similarly, f(3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

    f(4) = f(1 + 3) = f (1) f (3) = 3 × 27 = 81

    f(1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P.

    First term is 3 and common ratio is also 3. 

    Also we know 

    {tex}=> S_n = {a(r^n -1)\over r - 1}{/tex}       

    {tex}=> S_n = {3(3^n -1)\over 3 - 1}{/tex}  .................... (3)

    From (2) and (3), We get 

    {tex}=> 120 = {3(3^n -1)\over 3 - 1}{/tex}

    {tex}=> 120 = {3(3^n -1)\over 2}{/tex}

    {tex}=> 80 = 3^n -1{/tex}

    {tex}=> 81 = 3^n {/tex}

    {tex}=> 3^4 = 3^n {/tex}

    n = 4,

    So Value of n = 4

    Answered by Naveen Sharma (Mar 31, 2017 4:53 p.m.)
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please what is the  meaning of general and pricipal solution in trignometry.

Posted by Racheal Asamoah (Mar 29, 2017 8:26 a.m.) (Question ID: 4580)

4 cards are drawn at a time from a pack of 52 cards find the probability of getting all the 4 cards of the same suit. 

 

Posted by Tanu Shree (Mar 24, 2017 8:17 a.m.) (Question ID: 4474)

How to find the ratio  in which the line segment forming the points (4,8,10) and (6,10, -8) is divided by yz- plane

 

Posted by Sita Kadian (Mar 14, 2017 7:04 p.m.) (Question ID: 3782)

How to find the square root of -7-24i ?

Posted by Sita Kadian (Mar 14, 2017 6:59 p.m.) (Question ID: 3781)

  • Answers:
  • Ans. 

    Let {tex}\sqrt {-7-24i} = x+yi{/tex}

    Squaring Both Sides,

    => -7-24i = x2 - y+ 2xyi

    on comparing, we get

    => x2 - y= -7   ……(1)

    and 2xy = -24 ………(2)

    We know

    (x2+y2)= (x2-y2)2 + 4x2y2

    => (x2+y2)2 = 49 + 576

    => (x2+y2)= 625

    => x2+ y2 = 25  …………(3)

    solving (2) and (3), We get

    {tex}x = \pm 3 \space \space \space and \space y =\pm4{/tex}

    from (2) as product of xy is -ve, so it means x and y are of opposite sign.

    So, when x = 3, y = -4

    when x = -3, y = 4

    Therefore, 

    {tex}\sqrt {-7-24i} = \pm (3-4i){/tex}

    Answered by Naveen Sharma (Mar 16, 2017 9:42 a.m.)
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Find the term independent of x in the expansion (x/√3)+(√3/2x²)^10

 

Posted by Gurleen Kaur (Mar 11, 2017 9:04 p.m.) (Question ID: 3544)

  • your question is worng when i am solved then i find r=10/3 which is not natural

    corrwct question is {tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

    Posted by Neeraj Sharma (Mar 12, 2017 1:29 p.m.)
  • Please answer this question I need it urgently

     

    Posted by Gurleen Kaur (Mar 12, 2017 1:16 p.m.)
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  • Answers:
  • {tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

    {tex}{T_{r + 1}}{ = ^{10}}{C_r}{\left( {\sqrt {{x \over 3}} } \right)^{10 - r}}{\left( {{3 \over {2{x^2}}}} \right)^r}{/tex}

    {tex}{ = ^{10}}{C_r}{\left( x \right)^{5 - {r \over 2} - 2r}}{\left( 3 \right)^{ - 5 + {r \over 2} + r}}{\left( 2 \right)^{ - r}}{/tex}

    {tex}{ = ^{10}}{C_r}{\left( x \right)^{{{10 - 5r} \over 2}}}{\left( 3 \right)^{{{3r - 10} \over 2}}}{\left( 2 \right)^{ - r}}{/tex}

    {tex}for\,independent\,of\,x{/tex}

    {tex}{\left( x \right)^{{{10 - 5r} \over 2}}} = {x^0}{/tex}

    {tex}{{10 - 5r} \over 2} = 0{/tex}

    {tex}10 - 5r = 0{/tex}

    {tex}r = 2{/tex}

    {tex}independent\,term\,{T_3}{ = ^{10}}{C_2}{\left( 3 \right)^{ - 2}}{\left( 2 \right)^{ - 2}}{/tex}

    {tex} = {{10!} \over {2!8!}} \times {1 \over {36}}{/tex}

    {tex} = {{10 \times 9} \over {2 \times 36}}{/tex}

    {tex} = {5 \over 4}{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 1:47 p.m.)
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Find Mode-

20-24

24-29

30-34

35-39

40-44

Posted by Renu Gupta (Mar 07, 2017 10:23 p.m.) (Question ID: 3261)

  • What is the frequency?

    Posted by Manish Gandhi (Mar 08, 2017 12:48 p.m.)
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find the derivative of sin2x

Posted by Prem Sharma (Mar 06, 2017 9:39 p.m.) (Question ID: 3201)

  • Answers:
  • {tex}let\,y = \sin 2x{/tex}

    {tex}differentiating\,with\,respect\,to\,x{/tex}

    {tex}{{dy} \over {dx}} = {d \over {dx}}\left( {\sin 2x} \right){/tex}

    {tex} = \cos 2x{d \over {dx}}\left( {2x} \right){/tex}

    {tex} = 2\cos 2x{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 1:51 p.m.)
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Find derivative of function f(x)=sinx/sinx-cosx with respect to x.

Posted by Rudra Choudhary (Mar 06, 2017 9:03 p.m.) (Question ID: 3200)

  • Answers:
  • Ans. {tex}f(x) = {sinx\over sinx-cosx}{/tex}

    Using quotient Rule, we get

    {tex}f'(x) ={ {(sinx-cosx){d(sinx)\over dx}- sinx{d(sinx-cosx)\over dx}}\over (sinx-cosx)^2}{/tex}

    {tex}=> { (sinx-cosx)cosx - sinx(cosx-(-sinx)) \over (sinx-cosx)^2}{/tex}

    {tex}=> {sinx cosx - cos^2x -sinxcosx -sin^2x\over (sinx-cosx)^2}{/tex}

    {tex}=> {- (cos^2x +sin^2x)\over (sinx-cosx)^2}{/tex}

    {tex}=> {-1\over (sinx-cosx)^2}{/tex}

    Answered by Naveen Sharma (Mar 06, 2017 10:21 p.m.)
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I have a doubt in chapter 9 ie, sequences and series.

Question. insert 3 nos. b/w 1 & 256 so that the resulting sequence is a GP

  

Posted by JAHANVI MALHOTRA (Mar 05, 2017 6:58 p.m.) (Question ID: 3126)

  • anybody to answer this ques.

    Posted by JAHANVI MALHOTRA (Mar 05, 2017 7:48 p.m.)
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  • Answers:
  • {tex}let\,three\,numbers\,b/w\,1\,and\,256\,be\,{G_1},\,{G_2},\,and\,{G_3}{/tex}

    {tex}1,{G_1},\,{G_2},\,{G_3},256\,are\,in\,GP{/tex}

    {tex}a = 1{/tex}

    {tex}l = 256{/tex}

    {tex}a{r^4} = 256{/tex}

    {tex}{r^4} = 256{/tex}

    {tex}r = 4{/tex}

    {tex}Now\,{G_1} = ar = 1 \times 4 = 4{/tex}

    {tex}{G_2} = a{r^2} = 1 \times {\left( 4 \right)^2} = 16{/tex}

    {tex}{G_3} = a{r^3} = 1 \times {\left( 4 \right)^3} = 64{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 2:02 p.m.)
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find limits of limx→infinity (root a+x - root 9/x) 

Posted by tushar choudhary (Mar 05, 2017 2:36 p.m.) (Question ID: 3103)

  • please update your question. please wirte question correctly

    Posted by Neeraj Sharma (Mar 12, 2017 2:07 p.m.)
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let f:r-r be a function given by f(x)=x2+2.find f-1(27).

Posted by tushar choudhary (Mar 05, 2017 2:31 p.m.) (Question ID: 3102)

  • Answers:
  • {tex}let\,{f^{ - 1}}\left( {27} \right) = x{/tex}

    {tex} \Rightarrow f\left( x \right) = 27{/tex}

    {tex} \Rightarrow {x^2} + 2 = 27{/tex}

    {tex} \Rightarrow {x^2} = 25{/tex}

    {tex} \Rightarrow x = \pm 5{/tex}

    {tex}Hence{/tex}

    {tex}{f^{ - 1}}\left( {27} \right) = \left\{ { - 5,5} \right\}{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 2:16 p.m.)
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find two no. such that their arithematic mean is 5 and G.M is 9 without using the identity (a+b)2=(a-b)2+4ab

Posted by tushar choudhary (Mar 05, 2017 2:28 p.m.) (Question ID: 3100)

  • Answers:
  • There are no two numbers such that their arithmatic mean is 5 and G.M. is 9

    Because we have know that 

    AM>GM for between any two numbers a and  b.

    here AM=5 and GM=9

    9>5 or GM>AM which is invalid

    Answered by Neeraj Sharma (Mar 12, 2017 2:36 p.m.)
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If tan=1/3 and tan b=1/2 . prove that sin2(a+b)=1

Posted by tushar choudhary (Mar 05, 2017 2:25 p.m.) (Question ID: 3099)

  • Answers:
  • {tex}Given\,that{/tex}

    {tex}\tan A = {1 \over 3}\,and\,\tan B = {1 \over 2}{/tex}

    {tex}Now\,\,\tan \left( {A + B} \right) = {{\tan A + \tan B} \over {1 - tan A \cdot \tan B}}{/tex}

    {tex} = {{\left( {{1 \over 3} + {1 \over 2}} \right)} \over {\left( {1 - {1 \over 3} \cdot {1 \over 2}} \right)}}{/tex}

    {tex} = {{\left( {{5 \over 6}} \right)} \over {\left( {{5 \over 6}} \right)}}{/tex}

    {tex}\tan \left( {A + B} \right) = 1{/tex}

    {tex}A + B = {45^ \circ }{/tex}

    {tex}Now\,Sin2\left( {A + B} \right) = \sin \left( {2 \times {{45}^ \circ }} \right){/tex}

    {tex} = \sin {90^ \circ }{/tex}

    {tex}\sin 2\left( {A + B} \right) = 1{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 3:50 p.m.)
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If f(x) is a function that contain 3 times column and range and satisfy the relation. f(f(x)).(1+f(x)) = -f(x).

help me please fast

 

Posted by tushar choudhary (Mar 05, 2017 2:23 p.m.) (Question ID: 3098)

If A.M and G.M of two positive number a and b are 10 and 8 respectively find the number

Posted by vivek Joshi (Mar 05, 2017 9:46 a.m.) (Question ID: 3084)

  • Answers:
  • {tex}let\,two\,numbers\,be\,a\,and\,b{/tex}

    {tex}A.M. = {{a + b} \over 2} = 10{/tex}

    {tex}a + b = 20 \ldots \ldots \left( 1 \right){/tex}

    {tex}G.M. = \sqrt {ab} = 8{/tex}

    {tex}ab = 64 \ldots \ldots \left( 2 \right){/tex}

    {tex}Now\,{\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab{/tex}

    {tex} = {\left( {20} \right)^2} - 4 \times 64{/tex}

    {tex} = 400 - 256{/tex}

    {tex}{\left( {a - b} \right)^2} = 144{/tex}

    {tex}a - b = \pm 12 \ldots \ldots \left( 3 \right){/tex}

    {tex}solving\,\left( 1 \right)\,and\,\left( 3 \right){/tex}

    {tex}a = 4,b = 16\,or\,a = 16,b = 4{/tex}

    {tex}hence\,two\,numbers\,are\,4\,and\,16\,or\,16\,and\,4.{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 4:08 p.m.)
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A and B are two mutually exclusive and exhaustive events of a random experiment such that P (A)=6 [P (B)]^2 where P  (A)andP (B ) denotes probilities of A and B respectively find P (A) andP (B)

Posted by Randeep Singh (Mar 03, 2017 9:33 a.m.) (Question ID: 2948)

  • Answers:
  • {tex}let\,P\left( B \right) = x{/tex}

    {tex}then\,P\left( A \right) = 6{\left[ {P\left( B \right)} \right]^2} = 6{x^2}{/tex}

    {tex}Since\,A\,and\,B\,are\,mutually\,exclusive\,and\,exhaustive\,events{/tex}

    {tex}therefore{/tex}

    {tex}A \cup B = S{/tex}

    {tex}P\left( {A \cup B} \right) = P\left( S \right){/tex}

    {tex}P\left( {A \cup B} \right) = 1{/tex}

    {tex}P\left( A \right) + P\left( B \right) = 1{/tex}

    {tex}6{x^2} + x = 1{/tex}

    {tex}6{x^2} + x - 1 = 0{/tex}

    {tex}\left( {2x + 3} \right)\left( {3x - 1} \right) = 0{/tex}

    {tex}x = - {2 \over 3},{1 \over 3}{/tex}

    {tex}hence{/tex}

    {tex}P\left( B \right) = {1 \over 3}{/tex}

    {tex}P\left( A \right) = 6 \times {\left( {{1 \over 3}} \right)^2} = {2 \over 3}{/tex}

    Answered by Neeraj Sharma (Mar 12, 2017 4:26 p.m.)
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Differentiate Sinx with respect to x from 1st principle method

 

Posted by Innocent Geeta (Mar 03, 2017 1:06 a.m.) (Question ID: 2940)

  • Answers:
  • Ans.  By First Principle  \(f'(x) = lim_{h\to 0} \space{ f(x+h) - f(x) \over h}\)

    \(=> lim_{h\to 0} \space{ sin^2(x+h) - sin^2x \over h}\)

    \(=> lim_{h\to 0} \space{ [sin(x+h) + sinx]\times [sin(x+h) - sinx]\over h}\)      \([Using \space (a^2 - b^2) = (a+b)(a-b)]\)

    \(=> lim_{h\to 0} \space{2 sin({x+h+x\over 2}) cos({x+h-x\over 2})\times 2cos({x+h+x\over 2}) sin({x+h-x\over 2})\over h}\)

      \([Using \space sin a + sin b = 2 sin({a+b\over 2})cos ({a-b\over 2})\) and \( sin a - sin b = 2 cos({a+b\over 2})sin ({a-b\over 2})]\)

    \(=> lim_{h\to 0} \space4{ sin({2x+h\over 2}) cos{h\over 2} \space cos({2x+h\over 2}) sin{h\over 2}\over h}\)

    \(=> \space4 sin({2x+0\over 2}) cos{0\over 2} \space cos({2x+0\over 2}) lim_{h\to 0} {sin{h\over 2}\over {2h\over 2}}\)

    \(=> 4 sin x. cos 0. cos x. lim_{{h\over 2}\to 0} {1\over 2}{sin{h\over 2}\over {h\over 2}}\)        \([as \space h \to 0, then, {h\over 2}\to 0 ]\)

    \(=> 4 sin x. cos x. {1\over 2}\)       \([Using \space Identity, lim_{x \to 0 } \space {sinx \over x } = 1]\)

    \(=> 2 sin x. cos x\)

    Answered by Naveen Sharma (Mar 03, 2017 1:33 p.m.)
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A bag contains 100 tickets numbered  1 to 100 a number is drawn randomly .Find the probability that it is multiple of 5 or 7.

Posted by Gurleen Kaur (Mar 01, 2017 12:22 p.m.) (Question ID: 2855)

  • Answers:
  • Total no. of tickets = 100

    No. of ticets which are multiple of 5 = 20

    NO. of tickets which are multiple of 7 = 14

    NO.of tickets which are multiple of both i.e. multiple of 35:- 2

    Therefore tickets which are multiple of 5 or 7 :-  20+14-2 = 32

    Therefore probability :-   32/100 = 0.32

    Answered by Devanshu Agarwal (Mar 01, 2017 12:34 p.m.)
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If tan a=1÷3,tan b=1÷2 then prove that

Sin2(a+b)=1

Posted by Gurleen Kaur (Mar 01, 2017 10:15 a.m.) (Question ID: 2847)

  • Yes

    Posted by Gurleen Kaur (Mar 09, 2017 11:33 a.m.)
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  • Answers:
  • is this a correct question

    Answered by Innocent Geeta (Mar 02, 2017 9:50 p.m.)
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Write the domain of function of f(x)=

[X^2-2x+3]÷[x^2- x-20]

Posted by Gurleen Kaur (Feb 28, 2017 4:27 p.m.) (Question ID: 2799)

  • Answers:
  • f(x) = \((x^2 - 2x +3)/((x+4)(x-5))\)

    So Since denominator should not be zero therefore x should not be -4 and 5

    Hence domain x belongs to R - {-4,5}

    Answered by Devanshu Agarwal (Mar 01, 2017 10:01 a.m.)
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Find the term independent of x in the expansion of (x÷√3+√3÷2x^2)

Posted by Gurleen Kaur (Feb 28, 2017 4:23 p.m.) (Question ID: 2798)

  • Please answer

     

    Posted by Gurleen Kaur (Mar 01, 2017 12:18 p.m.)
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Tan a=1÷3, tan b=1÷2 prove that sin2(a+b)=1

Posted by Gurleen Kaur (Feb 28, 2017 4:20 p.m.) (Question ID: 2797)

  • Answers:
  • sin 2(a + b) = 1

    2 tan(a + b)/{1 - tan2(a + b)} = 1

    2[(tan a + tan b)/(1 - tana.tanb)]/[1 + {(tan a + tan b)/(1 - tana.tanb)}2] = 1

    2{(1/3 + 1/2)/1 - 1/3 x 1/2)]/[1 +{(1/3 + 1/2)/(1 - tan a .tan b)}2] = 1

    2[(5/6)/(5/6)]/[1 + {(5/6)/(5/6)}2] = 1

    2/[1 + 1] = 1

    2/2 = 1

    1 = 1

    Hence proved.

     

     

    Answered by Rashmi Bajpayee (Feb 28, 2017 7:14 p.m.)
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Tan32+tan13+tan32tan13=1

Posted by Gurleen Kaur (Feb 27, 2017 4:58 p.m.) (Question ID: 2731)

  • Answers:
  • tan 32o + tan 13o + tan 32o.tan13o = 1

    tan 32<font size="2">o</font> + tan 13<font size="2">o</font> = 1 - tan 32<font size="2">o</font>.tan13<font size="2">o</font>

    (tan 32<font size="2">o</font> + tan 13<font size="2">o)/(</font>1 - tan 32<font size="2">o</font>.tan13<font size="2">o</font><font size="2">) = 1</font>

    <font size="2">tan (32o + 13o) = 1                                        [Since tan (A + B) = (tan A + tan B)/(1 - tan A. tan B)]</font>

    <font size="2">tan 45o = 1</font>

    <font size="2">1 = 1                                                            [Since tan 45o = 1]</font>

    <font size="2">Hence proved.</font>

    Answered by Rashmi Bajpayee (Feb 27, 2017 7:11 p.m.)
    Thanks (1)
  • Hello friends,

    If we can recall the formula

    tan( A+B)= ( tan A + tan B)  /(1- tan A tan B)

    so let us take

    A= 32 and B= 13

    therefore   tan( 32+13) = (tan 32+ tan 13)/(1-tan32 tan13)

    i.e    tan 45 = (tan 32+ tan 13)/(1-tan32 tan13)        [since 32+13 =45]

     1= (tan 32+ tan 13)/(1-tan32 tan13)       [tan 45 =1]

    (1-tan32 tan13) = (tan 32+ tan 13)

    1=  tan32 tan13 + tan 32+ tan 13

    PROOVED

    Thanks

    S Mukherjee

    7864927899

     

    Answered by Twentyfirst Learning (Feb 27, 2017 6:43 p.m.)
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Sin 780°.Sin120°+Cos240°.Sin390°=1/2

Posted by Roma Lohia (Feb 25, 2017 11:26 a.m.) (Question ID: 2596)

  • Answers:
  • Ans. \(Sin 780°.Sin120°+Cos240°.Sin390°={1\over 2}\)

    Taking LHS

    \(Sin 780°.Sin120°+Cos240°.Sin390°\)

    => \(Sin( 2\times 360°+60°).Sin(180°-60°)+Cos(180°+60°).Sin(360°+30°)\)

    \(=> Sin( 4\pi+60°).Sin(\pi-60°)+Cos(\pi+60°).Sin(2\pi+30°)\)

    => \(Sin60°.Sin60°- Cos60°.Sin30°\)

    => \({\sqrt 3\over 2}.{\sqrt 3\over 2} - {1\over 2}.{1\over 2} = {3\over 4} - {1\over 4} \)

    => \({2\over 4 } = {1\over 2 } = RHS \)

    Hence Proved

    Answered by Naveen Sharma (Feb 25, 2017 12:28 p.m.)
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