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1

### JEE Main 2013 (Offline)

The expression $${{\tan {\rm A}} \over {1 - \cot {\rm A}}} + {{\cot {\rm A}} \over {1 - \tan {\rm A}}}$$ can be written as:
A
$$\sin {\rm A}\,\cos {\rm A} + 1$$
B
$$\,\sec {\rm A}\,\cos ec{\rm A} + 1$$
C
$$\tan {\rm A} + \cot {\rm A}$$
D
$$\sec {\rm A} + \cos ec{\rm A}$$

## Explanation

Given expression can be written as

$${{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}$$

(As $$\tan A = {{\sin A} \over {\cos A}}$$ and $$\cot A = {{\cos A} \over {\sin A}}$$ )

$$= {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}$$

$$= {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}$$

$$= 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$$
2

### AIEEE 2012

In a $$\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}$$ If $$3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6$$ and $$4\sin Q + 3\cos P = 1,$$ then the angle R is equal to :
A
$${{5\pi } \over 6}$$
B
$${{\pi } \over 6}$$
C
$${{\pi } \over 4}$$
D
$${{3\pi } \over 4}$$

## Explanation

Given $$3$$ $$\sin \,P + 4\cos Q = 6$$ $$\,\,\,\,\,\,\,\,...\left( i \right)$$

$$4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Squaring and adding $$(i)$$ & $$(ii)$$ we get

$$9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 16\,{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$= 36 + 1 = 37$$

$$\Rightarrow 9\left( {{{\sin }^2}p + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}q} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37$$

$$\Rightarrow 9 + 16 + 24\sin \left( {P + Q} \right) = 37$$

[ As $${\sin ^2}\theta + {\cos ^2}\theta = 1$$ and

$$\sin A\cos B + \cos A\sin B$$ $$= \sin \left( {A + B} \right)$$ ]

$$\Rightarrow \sin \left( {P + Q} \right) = {1 \over 2}$$

$$\Rightarrow P + Q = {\pi \over 6}$$ or $${{5\pi } \over 6}$$

$$\Rightarrow R = {{5\pi } \over 6}$$ or $${\pi \over 6}$$

(as $$P + Q + R = \pi$$ )

If $$R = {{5\pi } \over 6}$$ then $$0 < P,Q < {\pi \over 6}$$

$$\Rightarrow \cos Q < 1$$ and $$\sin P < {1 \over 2}$$

$$\Rightarrow 3\sin P + 4\cos Q < {{11} \over 2}$$ which is not true.

So $$R = {\pi \over 6}$$
3

### AIEEE 2011

If $$A = {\sin ^2}x + {\cos ^4}x,$$ then for all real $$x$$:
A
$${{13} \over {16}} \le A \le 1$$
B
$$1 \le A \le 2$$
C
$${3 \over 4} \le A \le {{13} \over {16}}$$
D
$${{3} \over {4}} \le A \le 1$$

## Explanation

$$A = {\sin ^2}x + {\cos ^4}x$$

$$= {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)$$

$$= {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}$$

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

Now $$0 \le {\sin ^2}\left( {2x} \right) \le 1$$

$$\Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}$$

$$\Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}$$

$$\Rightarrow 1 \ge A \ge {3 \over 4}$$
4

### AIEEE 2010

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

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

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

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

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

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