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### AIEEE 2010

MCQ (Single Correct Answer)
Let $$\cos \left( {\alpha + \beta } \right) = {4 \over 5}$$ and $$\sin \,\,\,\left( {\alpha - \beta } \right) = {5 \over {13}},$$ where $$0 \le \alpha ,\,\beta \le {\pi \over 4}.$$
Then $$tan\,2\alpha$$ =
A
$${56 \over 33}$$
B
$${19 \over 12}$$
C
$${20 \over 7}$$
D
$${25 \over 16}$$

## Explanation

$$\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}$$

$$\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}$$

$$\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$$

$$= {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}$$
2

### AIEEE 2009

MCQ (Single Correct Answer)
Let A and B denote the statements

A: $$\cos \alpha + \cos \beta + \cos \gamma = 0$$

B: $$\sin \alpha + \sin \beta + \sin \gamma = 0$$

If $$\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - {3 \over 2},$$ then:

A
A is false and B is true
B
both A and B are true
C
both A and B are false
D
A is true and B is false

## Explanation

3

### AIEEE 2006

MCQ (Single Correct Answer)
The number of values of $$x$$ in the interval $$\left[ {0,3\pi } \right]\,$$ satisfying the equation $$2{\sin ^2}x + 5\sin x - 3 = 0$$ is
A
4
B
6
C
1
D
2

## Explanation

$$2{\sin ^2}x + 5\sin x - 3 = 0$$

$$\Rightarrow \left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0$$

$$\sin x = {1 \over 2}$$ and $$\,\,\sin x \ne - 3$$

Given that $$x \in \left[ {0,3\pi } \right]$$

So possible values of x are $$30^\circ$$, $$150^\circ$$, $$390^\circ$$, $$510^\circ$$. That means x have 4 values.
4

### AIEEE 2006

MCQ (Single Correct Answer)
If $$0 < x < \pi$$ and $$\cos x + \sin x = {1 \over 2},$$ then $$\tan x$$ is
A
$${{\left( {1 - \sqrt 7 } \right)} \over 4}$$
B
$${{\left( {4 - \sqrt 7 } \right)} \over 3}$$
C
$$- {{\left( {4 + \sqrt 7 } \right)} \over 3}$$
D
$${{\left( {1 + \sqrt 7 } \right)} \over 4}$$

## Explanation

$$\cos x + \sin x = {1 \over 2}$$

$$\Rightarrow {\left( {\cos x + {\mathop{\rm sinx}\nolimits} } \right)^2} = {1 \over 4}$$

$$\Rightarrow {\cos ^2}x + {\sin ^2}x + 2\cos x\sin x = {1 \over 4}$$
$$\left[ \because {{{\cos }^2}x + {{\sin }^2}x = 1\, \,and \,\,2\cos x\sin x = \sin 2x} \right]$$

$$\Rightarrow 1 + \sin 2x = {1 \over 4}$$

$$\Rightarrow \sin 2x = - {3 \over 4},$$ so $$x$$ is obtuse and

$${{2\tan x} \over {1 + {{\tan }^2}x}} = - {3 \over 4}$$

$$\Rightarrow 3{\tan ^2}x + 8\tan x + 3 = 0$$

$$\therefore$$ $$\tan x = {{ - 8 \pm \sqrt {64 - 36} } \over 6}$$

$$= {{ - 4 \pm \sqrt 7 } \over 3}$$

as $$\tan x < 0\,$$

$$\therefore$$ $$\tan x = {{ - 4 - \sqrt 7 } \over 3}$$

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