Solution:

`f(x)=\tan^{-1}\ (\sqrt {\ frac {1+\sin x}{1-\sin x}}) x \in \ (0, \ frac {\pi}{2}\ )` ...Given

`f(x)=\tan^{-1}\ (\sqrt {\ frac {1+\sin x}{1-\sin x}}) `

`= tan^{-1}\ (\sqrt {\ frac { cos^2(x/2)+ sin^2(x/2) +2 sin (x/2)cos(x/2)}{ cos^2(x/2)+ sin^2(x/2) - 2sin (x/2)cos(x/2)}})`

` = \tan^{-1}\ (\sqrt ( (cos(x/2) + sin(x/2))^2/(cos(x/2) - sin(x/2))^2) ) `

` = tan^(-1) \( (1 + tan(x/2))/(1 - tan(x/2)))`

` = tan^(-1) \ \ (tan (pi/4) + tan(x/2))/(1 - tan(pi/4 ) tan(x/2))`

` = tan^(-1) tan(pi/4 + x/2 )`

` = pi/4 + x/2`

`f'(x) = 1/2`


`\therefore` Slope of normal `= -2`

So, equation of line is `\ frac {y-\frac {\pi}{3}}{x-\frac {\pi}{6}}=-2`

`(0,\ frac {2\pi}{3})` satisfies this equation `\ frac {\frac {-\pi}{3}}{\frac {\pi}{6}}=-2`

So, `(0,\frac {2\pi}{3})` lies on the line.

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Solution:

Let `tan^{-1}x = \theta`

Hence, `tan^{-1} ( frac{2x}{1-x^2} ) = 2 \theta`

`tan^{-1}y=tan^{-1}x+tan^{-1} (\frac {2x}{1-x^2} )`

`tan^(-1) y = theta + 2 theta = 3 theta = 3 tan^(-1)x `

Lets check for Range of `3tan^(-1)x`

`|x| < \ frac{1}{\sqrt 3}`

Hence, `\ frac {-\pi}{6} < \theta < \ frac{\pi}{6}`

Hence, `\ frac{-\pi}{2} < 3 \theta < \ frac{\pi}{2}` (setisfy the range of `tan^(-1)` function)

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Solution:

`\alpha = 3 \sin^{-1} \frac{6}{11} > 3\sin^{-1} \frac{6}{12}> 3\sin^{-1} \frac{1}{2} > \frac{\pi}{2}`

`\beta = 3 \cos^{-1} \frac{4}{9} > 3\cos^{-1} \frac{4}{8}> 3\cos^{-1} \frac{1}{2} >\pi`

`\therefore \alpha+\beta > \ frac{3\pi}{2}`

Hence `\sin \beta < 0, \cos\alpha < 0, \cos(\alpha+\beta) > 0`

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Solution:

` f(x)=2 tan^{-1}x + sin^{-1} ( \frac{2x}{1 + x^2} )`

For `x > 1`

`\sin ^{ -1 } ( \frac { 2x }{ 1+ x ^{ 2 } } ) =\pi-2\tan^{-1}x`

`\Rightarrow f(x) =2\tan^{-1}x+\pi-2\tan^{-1}x`

`\Rightarrow f(x)=\pi `

`\Rightarrow f(5)=\pi`

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Solution:

`tan^{-1}(cot \frac {3\pi}{4})=tan^{ -1 }[cot (\frac { \pi }{ 2 } +\frac { \pi }{ 4 } ) ]`

`=tan^{ -1 }[-\tan \frac { \pi }{ 4 } ] `
.
`=tan^{ -1 }[\tan (-\frac { \pi }{ 4 } ) ]`

` =-\frac { \pi }{ 4 }`

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Solution:

`f:[0, 4 \pi] \rightarrow [0, \pi ], f(x) = cos^{-1} (cos x)`

`\Rightarrow oint A, B, C` satisfy `f (x) = \frac{10-x}{10}`

Hence, 3 points

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Solution:

`\cot(\sum_{n=1}^{23}\cot^{-1}(1+\sum_{k=1}^{n}2k))`

= ` \cot(\sum_{n=1}^{23}\cot^{-1}(n^{2}+n+1))`

= ` \cot (\sum _{ n=1 }^{ 23 } \tan ^{ -1 } (\frac { n+1-n }{ 1+n(n+1) } )) =\cot ( \sum _{ n=1 }^{ 23} \tan ^{ -1 } ( n+1 ) -\tan ^{ -1 } n ) =\cot ( \tan ^{ -1 }24 -\tan ^{ -1 }1 ) `

=`\cot ( \tan ^{ -1 } ( \frac { 23 }{ 25 } ) ) `

= `\cot(\cot^{-1}(\frac{25}{23}))= \frac{25}{23}`

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Solution:

`x,y,z` are in `A.P`.

`=>2y=x+zcdots`(i)

Also `tan^-1x,tan^-1y,tan^-1z` are in `A.P`.

`=>2tan^-1y=tan^-1x+tan^-1z`

`=>tan^-1((2y)/(1-y^2))=tan^-1((x+z)/(1-xz))`

`=>(2y)/(1-y^2)=(x+z)/(1-xz)`

`=>(x+z)/(1-y^2)=(x+z)/(1-xz)` from (i)

`=>1-y^2=1-xz`

`=>y^2=xz`

`=>(x+z)^2=4xz`

`=>(x-z)^2=0`

Hence `x=y=z`

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Solution:

Here, innermost function is inverse.

`:. ` Put `tan^-1 y = theta => tan theta = y`

`[ 1/y^2 { ( cos ( tan^-1 y ) + y sin ( tan^-1 y ) )/ ( cot ( sin^-1 y ) + tan ( sin^-1 y ) ) }^2 + y^4 ] ^(1//2)`

Now `tantheta = (y/1) = "hieght"/"Base"`

`"hypotenuse" = sqrt("hieght"^2 + "Base"^2)`

` = sqrt(1+y^2)`

`cos theta = "Base" /"hypotenuse" = 1 / sqrt(1+y^2)`

`theta = tan^(-1)y = cos^(-1)(1/sqrt(1+y^2)) = sin^(-1) (y/sqrt(1+y^2))`

By putting these in equation

`= [ 1/y^2{ (1/sqrt(1 +y^2) + y^2/ sqrt(1 +y^2))/(sqrt(1 -y^2)/y + y/sqrt( 1 -y^2) ) }^2 + y^4 ] ^(1//2)`

`= [ 1/y^2* y^2 ( 1- y^4 ) + y^4 ] ^(1//2) = 1`

Q.

Given, `cos x + cos y = - cos z`

and `sin x + sin y = - sin z`

On squaring and adding, we get

`cos^2 x + sin^2 x + cos^2 y + sin^2 y + 2 cos x cos y + 2 sinx siny = 1`

`=> 2 + 2 [ cos ( x -y ) ] = 1 => cos ( x -y ) = 1/2 `

`=> 2 cos^2((x -y)/2 ) -1 = -1/2`

`=> 2 cos ^2 ((x -y ) /2 ) = 1/2`

`=> cos ((x -y)/2) = 1/2`

R .

`cos 2 x [ cos ( pi/4 - x ) - cos ( pi/4 + x ) ] + 2 sin^2 x = 2 sin x* cos x`

`=> cos 2 x ( sqrt2 sin x ) + 2 sin62 x = 2 sin x * cos x`

`=> sqrt2 sin x [ cos 2x + sqrt2 sin x - sqrt 2 cos x ] = 0`

`=> sin x = 0 , ( cos x - sin x ) ( cos x + si x - sqrt2) = 0`

`=> sec x =1 ` or `tan x =1 `

`=> secx = 1 ` or `sqrt2`

S.

`cot ( sin^-1 sqrt ( 1 -x^2 ) = sin( tan^-1 ( x sqrt 6 ))`

`=> x/sqrt(1 -x^2 ) = (x sqrt6)/(sqrt (1 + 6x^2 ))`

`=> 1 + 6 x^2 = 6 - 6x^2`

`=> 12 x^2 = 5 => x = sqrt(5/12 ) = sqrt5/ (2 sqrt3 )`

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Solution:

Let `cot^(-1)x=y=>x=coty..`..(1).

Let `cos^-1p=cot^(-1)x`

`=> cos^(-1)p=y`

`=>p=cosy.(Let 0<=p<=1)`

`cosy/siny=coty=>cosy=coty/cscy, ["cosec"^2y = 1+cot^2y]`

`=>cosy=coty/(sqrt(1+cot^2y))....`.(2)

From equation (1),

`=>p=x/((1+x^2)^(1/2)).`

`siny = tanyxx cosy= cosy/coty`

From equation (1) and (2), we get

`=>siny = coty/(sqrt(1+cot^2y))xx 1/coty= 1/sqrt (1+x^2)`

Using these two results,

`sqrt(1+x^2)[{xcos(cot^(-1)x)+sin(cot^(-1)x)}^2-1]^(1/2)`

`=sqrt(1+x^2)[(x^2/((1+x^2)^(1/2))+1/((1+x^2)^(1/2)))^2-1]^(1/2)`

`=sqrt(1+x^2)(x^2+1-1)^(1/2)=xsqrt(1+x^2)`

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Solution:

Given that,

`sin^(-1) (x/5)+"cosec"^(-1)(5/4)=pi/2`

`=>sin^(-1) (x/5)+sin^(-1)(4/5)=pi/2`

`=>sin^(-1)(x/5)=pi/2-sin^(-1) (4/5)`

`=>sin^(-1)(x/5)=cos^(-1) (4/5)`

`=>sin^(-1) (x/5)= sin^(-1)(3/5)`

`=>x/5=3/5`

`=>x=3.`

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Solution:

`sin [cot^(-1) (1 + x)] = cos (tan^(-1)x) ...(i)` and we know,

`cot^(-1) θ = sin^(−1) (1/sqrt(1+θ^2)),`

and, `tan^(-1) θ = cos^(−1) (1/sqrt(1+θ^2))`

`∴` From Eq.` (i),`

`sin (sin^(−1) (1/sqrt(1 + (1+x)^2 ))) = cos(cos^(−1) (1/(1+x^2)))`

`⇒ 1/sqrt(1+(1+x)^2) = 1/sqrt(1+x^2)`

`⇒ 1 + x^2 + 2x + 1`

` = x^2 + 1`

`⇒ x = −1/2`

The correct answer is: `−1/2`

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Solution:

`f(x) = sqrt( sin^-1 2(x)+π/6)`,

to find domain we must have;

`sin^-1(2x)+π/6 ≥ 0 ( but−π/2 ≤ sin^-1 θ ≤ π/2)`

`−π/6 ≤ sin^-1(2x) ≤ π/2`

`⇒ sin(-π/6) ≤ 2x ≤ sin(π/2) ⇒ −1/2 ≤ 2x ≤ 1`

`⇒ −1/4 ≤ x ≤ 1/2 ⇒ x∈[-1/4,1/2]`

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Solution:

LHS `=cos tan^-1 [sin (cot^-1 x)]`

`= cos tan^-1 [ sin ( sin^-1 \ \ 1/ sqrt(1 +x^2) ) ]`

`= cos ( tan^-1 \ \ 1/sqrt ( 1 + x^2) ) = sqrt((x^2 + 1)/(x^2 + 2)) = ` RHS

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Solution:

We know that, `sin^(-1) (α) + cos^(-1) (α) = π/2`

Therefore, `α` should be equal in both functions.

`∴ x − x^2/2 + x^3/4 −... = x^2−x^4/2 + x^6/4 −...`

`⇒ x/(1+ x/2) = x^2/(1 + x^2/2)`

`⇒ x/((2+x)/2) = x^2/((2+x^2)/2)`

`⇒( 2x)/(2 + x) = ( 2x^2)/(2 + x^2)`

`⇒ 2x(2 + x^2) = (2x^2) (2 + x)`

`⇒ 4x + 2x^3 = 4x^2 + 2x^3`

⇒ Either `x = 0` or `4 - 4x = 0`

`⇒ x = 0` or `x = 1`

`
∵ 0 < |x| < sqrt2,`

`∴ x = 1`

The correct answer is: `1`

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Solution:

We know that, `sin^(-1) (alpha ) + cos^(-1) (alpha ) = pi/2`

Therefore, `alpha` should be equal in both functions.

`:. x - x^2/2 + x^3/4 - ... = x^2 - x^4/2 + x^6/4 - ...`

` => x/( 1 + x/2) = x^2/(1 + x^2/2) = x^2/(1 + x^2/2) => x/((2 + x)/2) = x^2/((2 + x^2 )/2)`

` => (2x)/( 2 + x) = (2 x^2)/( 2 + x^2)`

`=> 2x (2 + x^2 ) = 2x^2 (2 + x)`

`=> 4x + 2x^3 = 4x^2 + 2x^3`

`=> x (4 + 2x^2 - 4x - 2x^2 ) = 0`

`=>` Either `x = 0` or `4 - 4x = 0`

`=> x = 0` or `x = 1`

` ∵ 0 < | x | < sqrt2`

`:. x = 1` and `x != 0`

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Solution:

Given function is `tan^(−1) sqrt(x(x+1) ) + sin^(−1)sqrt(x^2+x+1) = π/2`

Function is defined, if

`1. x (x + 1 ) ≥ 0 ∵` Domain of square root function.

`2. x^2 + x + 1 ≥ 0 ∵` Domain of square root function.

`3. sqrt(x^2+x+1) ≤ 1 ∵` Domain of sin-1 function.

From `(ii)` and `(iii)`

`0 ≤ x^2 + x + 1 ≤ 1 ∩ x^2 + x≥ 0`

`⇒ 0 ≤ x^2 + x + 1 ≤ 1 ∩ x^2 + x + 1 ≥ 1`

`⇒ x^2 + x + 1 = 1`

`⇒ x^2 + x = 0`

`⇒ x(x + 1) = 0`

`⇒ x = 0, x = -1`

The correct answer is: `2`

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Solution:

Given , `A = 2 tan -1 (2sqrt2 - 1)`

and `B = 3 sin^-1\ \ 1/3 + sin ^-1\ \ 3/5`

Here `A =2 tan^-1 ( 2sqrt2 -1 )`

`=2 tan^-1 ( 2 xx 1.414 -1 ) = 2 tan^-1 ( 1.828)`

`:. A > 2 tan^-1 sqrt3 = 2 * pi/3 = ( 2 pi)/3`

To find the value of B, we first say

`sin^-1\ \1/3 < sin^-1\ \1/2 = pi/6`

so that ` 0 < 3 sin^-1\ \ 1/3 < pi/2`

Now , ` 3 sin^-1\ \1/3 = sin^-1 ( 3* 1/3 - 4 * 1/27 )`

`= sin^-1 ( 23/27 ) = sin^-1 ( 0.851) < sin^-1 (sqrt3/2) = pi/3`

`sin^-1\ \3/5 = sin^-1 (0.6) < sin^-1 ( sqrt3/2) = pi/3`

`:. B < pi/3 + pi/3 = ( 2pi)/3`

Thus , `A > ( 2pi)/3`

and `B < ( 2pi)/3`

Hence , greater angle is A

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Solution:

`sin^(−1) (sin\ \ (2π)/3) = sin^(−1) [sin(π−π/3)]`

`= sin^(−1) (sin \ \π/3) = π/3`

The correct answer is: `π/3`

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Solution:

`tan[2tan^(−1) \ \1/5−π/4] = tan [tan^(−1 ) ((2xx1/5)/(1−1/25))−π/4]`

`=tan[tan^(−1)(5/12−π/4)]`

`= (tan [tan^(−1) \ \5/12]−tan \ \π/4)/(1 + tan[tan^(−1)\ \5/12]tan \ \π/4)`

`=( 5/12 − 1)/(1+5/12 . 1) = −7/17`

The correct answer is: `−7/17`

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Solution:

`tan[ cos^-1\ \ 4/5 + tan^-1\ \ 2/ 3 ]`

`= tan [ tan^-1\ \3/4 + tan^-1\ \2/3 ] \ \ \ \ \ \ \ \ [ because cos^-1\ \4/5 = tan^-1\ \3/4 ]`

`= tan [ tan^-1 (( 3/4 + 2/3 )/( 1 - 3/4* 2/3 ) ) ]`

`= tan[ tan^-1\ \ 17/6 ] = 17/6`

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Solution:

`θ = tan^(−1) sqrt((a(a+b+c))/(bc)) + tan^(−1)sqrt( (b(a+b+c))/(ca)) + tan^(−1)sqrt( (c(a+b+c))/(ab)).`

`[∵ tan^(−1)x+ tan^(−1) y+ tan^(−1) z=tan^(−1) ((x+y+z−xyz)/(1−xy−yz−zx))]`

`=tan^(−1) {(sqrt(a+b+c) (sqrta/(bc)) + (sqrtb/(ca)) +(sqrtc/(ab)) −(a+b+c) sqrt((a+b+c)/(abc)))/(1−(a+b+c)(1/a+1/b+1/c))}`

`= tan^(−1) {(sqrt((a+b+c)/(abc)) (a+b+c)−(a+b+c) sqrt((a+b+c)/(abc)))/(1−(((a+b+c)(ab+bc+ca))/(abc)) )}`

`⇒ θ =tan^(−1) 0`

`⇒ tanθ = 0`

The correct answer is: `0`

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Solution:

Let `f(x) = cos (2 cos^(-1) x + sin^(-1)x)`

` [∵ cos^(−1) x + sin^(−1)x = π/2]`

`=cos(cos^(−1) x+π/2)`

`= -sin(cos^(-1) x)`

`⇒ f(x) = − sin(sin^(−1) (sqrt(1−x^2)))`

`⇒ f(1/5) = − sin (sin^(−1) sqrt(1−1/5^2))`

`= − sin(sin^(−1) ((2sqrt6)/5) )= − (2sqrt6)/5`

The correct answer is: `− (2sqrt6)/5`

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Solution:

Given, `tan^-1 2x + tan^-1 3x = pi/4`

`=> tan^-1 ((2x + 3x)/(1- 6x^2)) = pi/4`

`=> (5x)/(1 -6x^2) =1`

`=> 6x^2 + 5 x -1 =0`

`=> ( x +1 ) ( 6 x -1) = 0`

`=> x =-1 ` or `1/6`

But x = -1 does not satisfy the given equation.

`:.` We take `x = 1/6`

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