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}

{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}

{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}

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

{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}

{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}

{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}

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 \( sin a - sin b = 2 cos({a+b\over 2})sin ({a-b\over 2})]\)

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

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

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

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

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

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

is this a correct question

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}

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.

tan 32^o + tan 13^o + tan 32^o.tan13^o = 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 (32^o + 13^o) = 1                                        [Since tan (A + B) = (tan A + tan B)/(1 - tan A. tan B)]</font>

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

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

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

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

OK thanks,....

Ans.You start by X = Y 

it means (Y-X)=  0 And you can't divide anything by zero.

so dividing both sides by (Y-X) is illegal.

Ans. \(let \space (a+ib) = i^{3\over 2}\)

Squaring Both Sides, We get 

\(=> a^2 +b^2.i^2 +2abi = i^3\)

\(=> a^2 -b^2 +2abi = -i \space \space \space \space \space \space \space \space \space \space [i^2 = -1]\)

On Comparing Both sides, We get 

\(a^2-b^2 = 0 \space \space \space \space \space \space \space and \space \space \space \space \space \space 2ab =-1\)

=> \(a^2 = b^2 \space \space \space ...(1) \space \space \space \space \space and \space \space a = {-1\over 2b} \space \space \space \space \space ... (2)\)

Put value of a in (1)

=> \(({-1\over 2b})^2 = b^2 => {1\over 4b^2} = b^2 => {1\over 4} = b^4 \)

=> \(b^2 = {1\over 2} => b = {1\over \sqrt 2}\)

Put value of b in (2), we get \(a = {-1\over 2 \times {1\over \sqrt 2}} => a = {-1\over \sqrt2}\)

So
=> \({-1\over \sqrt2}+{1\over \sqrt 2}i = i^{3\over 2}\) 

Ans. \(cos^4{\pi \over 8}+cos^4{3\pi \over 8}+cos^4{4\pi \over 8}+cos^4{7\pi \over 8} ={3\over 2}\)

Taking LHS,

\(cos^4{\pi \over 8}+cos^4{3\pi \over 8}+cos^4{4\pi \over 8}+cos^4{7\pi \over 8}\)

=> \(cos^4{\pi \over 8}+cos^4{3\pi \over 8}+cos^4[{{\pi} -{3\pi \over 8}}]+cos^4[{{\pi - {\pi \over 8}}}]\)

=> \(cos^4{\pi \over 8}+cos^4{3\pi \over 8}+({-cos{3\pi \over 8}})^4+({-cos{\pi \over 8}})^4\)

=> \(2[cos^4{\pi \over 8}+cos^4{3\pi \over 8}]\)

=> \(2[cos^4{\pi \over 8}+cos^4({\pi \over 2}-{\pi \over 8})]\)

=> \(2[cos^4{\pi \over 8}+sin^4{\pi \over 8}]\)

\(=> 2[(cos^2{\pi \over 8}+sin^2{\pi \over 8})^2 -2 cos^2{\pi \over 8}.sin^2{\pi \over 8}]\)

\(=> 2[(1)^2 -{1\over 2}(2 cos{\pi \over 8}.sin{\pi \over 8})^2]\)

\(=> 2[1 -{1\over 2}(sin {\pi \over 4})^2]\)

\(=> 2[1 -{1\over 2}\times {1\over 2}]\)

\(=> 2[1 -{1\over 4}] => 2 \times {3\over 4} = {3\over 2} = RHS \)

Hence Proved

Ans. To Find Range,

\(Let \space y = {1 \over 1-x^2}\)

\(=> {1-x^2} = {1\over y} \)

\(=> x^2 ={1-{1\over y}}\)

\(=> x^2 = {y-1\over y}\)

\(=> x= {\sqrt{y-1\over y}}\)

\(Now,\space this \space is \space defined \space for \space {y-1\over y} \geq 0 \space except \space y \neq 0\)

\(y \in (-\infty , 0) \cup [1, \infty)\)

So Domain is \((-\infty , 0) \cup [1, \infty)\)

 It is wrong because we can't write tan3x = tan2x + tanx

The and of this question is 

Tan(2x+x) = tan2

tan 3x = tan 2x + tan x

tan (2x + x) = tan 2x + tan x

(tan 2x + tan x)/(1 - tan 2x.tanx x) = tan 2x + tan x

1 - tan 2x.tanx = 1

tan 2x. tan x = 0

tan 2x = 0 and tan x = 0

tan 2x = tan 0^o and tan x = tan 0^o

2x = 0^{<font size="2">o</font>} and x = 0^{<font size="2">o</font>}

x = 0^{<font size="2">o</font>} and x = 0^{<font size="2">o</font>}

Ans. we know

\(sin^3x = sinx(sin^2x)\)

\(=>{ sin x (1-cos2x)\over2}\)

\(=> {1\over 2} [{sin x -sin x cos2x}]\)

\(=> {1\over 2} [{ sin x - {1\over 2} [sin (x+2x) +sin (x-2x)]}]\)

\(=> {1\over 2} [{ sin x - {1\over 2} [sin 3x -sin x}]]\)

\(=> {1\over 2}\times{1\over 2} [{2 sin x - sin 3x +sin x}]\)

\(=> {3sin x - sin 3x \over4}\)  (1)

Similarly,

\(=> sin^3({{2\pi\over 3} +x})= {3sin ({{2\pi\over 3 }+x })- sin (2\pi +3x )\over4}\)

\(=> {3[sin ({{2\pi\over 3 })cos x +sin xcos( {2\pi\over 3} })]- sin 3x\over4} \)

\(=> {3[{\sqrt3\over 2}cos x -{1\over 2}sin x)]- sin 3x\over4} \)

\(=> {3\sqrt3cos x -3sin x- 2sin 3x\over 8} \)    (2)

Similarly, 

\(=> sin^3({{4\pi\over 3} +x})= {3sin ({{4\pi\over 3 }+x })- sin (4\pi +3x )\over4}\)

\(=> {3[sin ({{4\pi\over 3 })cos x +sin xcos( {4\pi\over 3} })]- sin 3x\over4} \)

\(=> {3[(-{\sqrt3\over 2 })cos x +sin x({-1\over 2})]- sin 3x\over4} \)

\(=> {-3\sqrt3cos x -3sin x-2sin 3x\over 8} \)  (3)

From (1),(2) and (3)

\(=> sin^3{x}+ sin^3({{2\pi\over 3} +x}) + sin^3({{4\pi\over 3} +x})\)

\(=>{3sin x - sin3x \over 4}+ {3\sqrt3cos x -3sin x- 2sin 3x\over 8} + {-3\sqrt3cos x -3sin x- 2sin 3x\over 8} \)

\(=>{1\over 8}[{6sin x - 2sin3x + 3\sqrt3cos x -3sin x- 2sin 3x -3\sqrt3cos x -3sin x- 2sin 3x}] \)

\(=>{1\over 8}[ {- 6sin 3x}] = {-6\over 8} {sin 3x} \)

\(=> {-3\over 4}{sin 3x} = RHS \)

Hence Proved

Ans. \(\lim_{x \to 0} {tan x - sin x \over x^3}\)

=> \(\lim_{x \to 0} {{sin x\over cos x} - sin x \over x^3}\)

=> \(\lim_{x \to 0} {sin x({1\over cosx } - 1) \over x^3}\)

=> \(\lim_{x \to 0} {sin x(1 - cosx) \over x^3 . \space cos x }\)

=> \(\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{ 1 - cosx \over x^2 }}]\)

=> \(\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{ 2sin^2{x\over 2} \over x^2 }}]\)

=> \(2\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{ sin{x\over 2} \over x}.{ sin{x\over 2} \over x}}]\)

=> \(2\lim_{x \to 0} {[{1\over cos x}. {sin x \over x} .{1\over 2}{ sin{x\over 2} \over {x\over 2}}.{1\over 2}{ sin{x\over 2} \over {x\over 2}}}]\)

=> \(2[\lim_{x \to 0} {{1\over cos x}\times \lim_{x \to 0}{sin x \over x} \times \lim_{x \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}\times \lim_{x \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}}]\)   

=> \(2[\lim_{x \to 0} {{1\over cos x}\times \lim_{x \to 0}{sin x \over x} \times \lim_{{x\over 2} \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}\times \lim_{{x\over2} \to 0}{1\over 2}{ sin{x\over 2} \over {x\over 2}}}]\)     [as x tends to zero, x/2 also tends to zero]

=> \(2\times 1 \times 1 \times {1\over 2}\times {1\over 2} = {1\over 2}\)

Ans. 

\(Derivative \space of \space \sqrt [3] {tan x} \space by \space First \space Principle \space is \space given \space by : \)

\(=> f'(x) = lim_{h \to 0} \space {{ \sqrt [3]{tan(x+h)} - \sqrt [3] {tan x}} \over h}\)

\(=> lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h)\over cos(x+h)} - \sqrt [3] {sinx\over cosx }} \over h}\)

\(=> lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx \space cos(x+h) }} \over h {\sqrt [3]{cos(x+h) cosx} }}\)

\(=> lim_{h \to 0} {1\over {\sqrt [3]{cos(x+h) cosx} }} . \space lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx \space cos(x+h) }} \over h}\)

\(=> {1\over {\sqrt [3]{cos(x+0) cosx} }} . \space lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx \space cos(x+h) }} \over h}\)

\(=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0} \space {{ \sqrt [3]{sin(x+h) cos \space x} - \sqrt [3] {sinx\space cos(x+h)}}\over h} \times{ {[{(sin(x+h) cos \space x})^{2\over 3}}+{({sinx \space cos(x+h)})^{2\over3}}+\sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}]\over {[{(sin(x+h) cos \space x})^{2\over 3}}+{({sinx \space cos(x+h)})^{2\over3}}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}]}\)   

\(\)\(=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0}{ \space {{sin(x+h) cos \space x - sinx\space cos(x+h)}} \over {h [{(sin(x+h) cos \space x})^{2\over 3}+({sinx \space cos(x+h)})^{2\over3}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}}]}\)                       \([a^3-b^3 = (a-b)(a^2+b^2+ab)]\)

\(=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0}{ \space {{sin(x+h -x )}} \over {h [{(sin(x+h) cos \space x})^{2\over 3}+({sinx \space cos(x+h)})^{2\over3}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}}]}\)             \([sin (a-b) = sina.cosb - cos a. sin b ]\)

\(=> {1\over {\sqrt [3]{cos^2x } }}.\space lim_{h \to 0}{ \space {{sinh }} \over h} . \space lim_{h \to 0}{ { 1\over [{(sin(x+h) cos \space x})^{2\over 3}+({sinx \space cos(x+h)})^{2\over3}+ \sqrt [3]{sin(x+h)cosx. \space cox(x+h)sinx}]}}\)

\(=> {1\over cos^{2\over3}x} . 1. {1\over[(sin(x+0) cos \space x)^{2\over 3}+(sinx \space cos(x+0))^{2\over3}+ \sqrt [3]{sin(x+0)cosx. \space cox(x+0)sinx}]}\)

\(=> {1\over cos^{2\over3}x} . 1. {1\over[(sinx cos \space x)^{2\over 3}+(sinx \space cosx)^{2\over3}+ \sqrt [3]{sinx.cosx. \space cosx .sinx}]}\)

\(=> {1\over cos^{2\over3}x} . 1. {1\over[(sinx cos \space x)^{2\over 3}+(sinx \space cosx)^{2\over3}+ ({sinxcosx})^{2\over3}]}\)

\(=> {1\over cos^{2\over3}x} . {1\over3(sinx cos \space x)^{2\over 3}}\)

\(=> {1\over 3}{1\over cos^{4\over3}x. sin^{2\over 3}x}\)

\(=> {1\over 3}{{1\over cos^2x}\over ({ cos^{4\over3}x. sin^{2\over 3}x\over cos^2x})}\)

\(=> {1\over 3}{{sec^2x}\over ({ sin^{2\over 3}x\over cos^{2\over 3}x})}\)

\(=> {1\over 3}.{1\over { tan^{2\over 3}x}}.{sec^2x}\)

Total arrangements possible are 7P7 = 7! = 5040

There are 2 vowels & 5 cosonants => there is only one way when the given condition will not meet & that is when both these vowels will occupy the odd places. Now, the no. of ways it is possible = 4P2 * 5P5 = 12 * 120 = 1440

=> Total words which can be formed in the required fashion are 5040-1440 = 3600 (Answer)

The question is : Prove that tan 4x= 4 tanx(1-tan²x)/1-6tan2x+tan⁴x

L.H.S.

tan 4x = tan 2(2x)

[We know that tan 2x = 2 tan x / 1 - tan ² x]

= 2 tan 2x / 1 - tan² (2x)

[now putting tan 2x = 2 tan x / 1 - tan²x]

= 2[2 tan x/1-tan²x] / 1 - [2 tan x / 1 - tan²x] ²

=[4 tan x / 1 - tan ² x] / [1 - 4 tan ² x / (1 - tan² x)²]

=[4 tan x / 1 - tan ² x] / [ (1- tan ² x⁾² - 4 tan ² x / (1 - tan² x)²]

= 4 tan x (1 - tan ² x) / (1- tan ² x)² - 4 tan ² x

= 4 tan x (1 - tan ² x) / 1 - 2 tan² x +tan ⁴ x - 4tan² x

= 4 tan x (1 - tan ² x) / 1 - 6 tan ² x + tan ⁴ x

sin5x + sinx = 2sin3x cos2x 
cos5x - cosx = -2sin3x sin2x           ..... use factorisation formulae of trigonometry. 
So the fraction is 2sin3x(cos2x-1)/ - 2sin3xsin2x 
= (1- cos2x)/sin2x 
= 2(sin²x)/ 2sinxcosx 
=sinx/cosx 
=tanx.

Ans. If question is: log (log sin x^1/2)

then according to Chain rule

\({dy\over dx} = { 1\over log( sin x^{1\over 2})} \times { 1 \over sin (x)^{1\over2}} \times cos (x)^{1\over 2}\times {1\over2\sqrt x}\)