 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2015 (Offline)

Let $${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right),$$
where $$\left| x \right| < {1 \over {\sqrt 3 }}.$$ Then a value of $$y$$ is:
A
$${{3x - {x^3}} \over {1 + 3{x^2}}}$$
B
$${{3x + {x^3}} \over {1 + 3{x^2}}}$$
C
$${{3x - {x^3}} \over {1 - 3{x^2}}}$$
D
$${{3x + {x^3}} \over {1 - 3{x^2}}}$$

## Explanation

Given,

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

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

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

$$\therefore$$ $$\,\,\,$$ $${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {{{3x - {x^2}} \over {1 - 3{x^2}}}} \right)$$

$$\Rightarrow y = {{3x - {x^3}} \over {1 - 3{x^2}}}$$
2

### JEE Main 2013 (Offline)

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 $$x=y=z.$$
3

### AIEEE 2008

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,

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

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

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

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

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

$$= {6 \over {17}}$$
4

### AIEEE 2007

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

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