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1

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)
If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then
A
$$x=y=z$$
B
$$2x=3y=6z$$
C
$$6x=3y=2z$$
D
$$6x=4y=3z$$

Explanation

Given that, $$x,y,z\,\,$$ are in $$AP$$

So, $$\,\,\,$$ $$2y = x + y$$

Also given that,

$${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$$ and $$\,\,\,{\tan ^{ - 1}}z\,\,$$ are in $$AP$$

So, $$2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$$

$$ \Rightarrow {\tan ^{ - 1}}\left( {{{2y} \over {1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {{{x + 7} \over {1 - xz}}} \right)$$

$$ \Rightarrow {{2y} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$$

$$ \Rightarrow {{x + z} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$$

[ as $$\,\,\,\,$$ $$2y = x + z$$]

$$ \Rightarrow 1 - {y^2} = 1 - xz$$

$$ \Rightarrow $$ $${y^2} = xz$$

As we get $${y^2} = xz,$$ so, $$x,y,z$$ are in $$GP.$$

According to the question $$x,y,z$$ are $$AP.$$

$$x,y,z$$ both can be $$AP$$ as well as $$GP$$

when $$x=y=z.$$
2

AIEEE 2008

MCQ (Single Correct Answer)
The value of $$cot\left( {\cos e{c^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$ is
A
$${{6 \over 17}}$$
B
$${{3 \over 17}}$$
C
$${{4 \over 17}}$$
D
$${{5 \over 17}}$$

Explanation

Given,

$$Cot\left( {so{{\sec }^{ - 1}}{5 \over 3} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$

$$ = \cot \left( {{{\tan }^{ - 1}}{3 \over 4} + {{\tan }^{ - 1}}{2 \over 3}} \right)$$

$$ = cot\left( {{{\tan }^{ - 1}}{{{3 \over 4} + {2 \over 3}} \over {1 - {3 \over 4} - {2 \over 3}}}} \right)$$

$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{17} \over 6}} \right)} \right)$$

$$ = \cot \left( {{{\cot }^{ - 1}}{6 \over {17}}} \right)$$

$$ = {6 \over {17}}$$
3

AIEEE 2007

MCQ (Single Correct Answer)
If sin-1$$\left( {{x \over 5}} \right)$$ + cosec-1$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$, then the value of x is
A
4
B
5
C
1
D
3

Explanation

Given sin-1$$\left( {{x \over 5}} \right)$$ + cosec-1$$\left( {{5 \over 4}} \right)$$ = $${\pi \over 2}$$

$$ \Rightarrow $$ sin-1$$\left( {{x \over 5}} \right)$$ + sin-1$$\left( {{4 \over 5}} \right)$$ = $${\pi \over 2}$$

$$ \Rightarrow $$ sin-1$$\left( {{x \over 5}} \right)$$ = $${\pi \over 2}$$ - sin-1$$\left( {{4 \over 5}} \right)$$

$$ \Rightarrow $$ sin-1$$\left( {{x \over 5}} \right)$$ = cos-1$$\left( {{4 \over 5}} \right)$$

$$ \Rightarrow $$ $${x \over 5}$$ = sin(cos-1$$ {{4 \over 5}}$$)

$$ \Rightarrow $$ $${x \over 5}$$ = sin(sin-1$$ {{3 \over 5}}$$)

$$ \Rightarrow $$ $${x \over 5}$$ = $${3 \over 5}$$

$$ \Rightarrow $$ x = 3
4

AIEEE 2005

MCQ (Single Correct Answer)
If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha ,$$ then $$4{x^2} - 4xy\cos \alpha + {y^2}$$ is equal to
A
$$2\sin 2\alpha $$
B
$$4$$
C
$$4{\sin ^2}\alpha $$
D
$$-4{\sin ^2}\alpha $$

Explanation

As we know,

$${\cos ^{ - 1}}A - {\cos ^{ - 1}}B$$

$$ = {\cos ^{ - 1}}\left( {AB + \sqrt {1 - {A^2}} .\sqrt {1 - {B^2}} } \right)$$

Given, $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha $$

$$ \Rightarrow {\cos ^{ - 1}}\left( {x.{y \over 2} + \sqrt {1 - {x^2}} .\sqrt {1 - {{{y^2}} \over 4}} } \right) = \alpha $$

$$ \Rightarrow {{xy} \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{y{}^2} \over 4}} = \cos \,x$$

$$ \Rightarrow {\left( {\cos x - {{xy} \over 2}} \right)^2} = \left( {1 - {x^2}} \right)\left( {1 - {{{y^2}} \over 4}} \right)$$

$$ \Rightarrow {\cos ^2} + {{{x^2}{y^2}} \over 4} - 2.\cos x.{{xy} \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {x^2} - {{{y^2}} \over 4} + {{{x^2}{y^2}} \over 4}$$

$$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\,\cos x = 1 - {\cos ^2}x$$

$$ \Rightarrow {x^2} + {{{y^2}} \over 4} - xy\cos x = {\sin ^2}x$$

$$ \Rightarrow 4{x^2} + y{}^2 - 4xy\cos x = 4{\sin ^2}x$$

Questions Asked from Inverse Trigonometric Functions

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