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1

### JEE Main 2014 (Offline)

Let $$fk\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ where $$x \in R$$ and $$k \ge \,.$$
Then $${f_4}\left( x \right) - {f_6}\left( x \right)\,\,$$ equals
A
$${1 \over 4}$$
B
$${1 \over 12}$$
C
$${1 \over 6}$$
D
$${1 \over 3}$$

## Explanation

Let $${f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)$$

Consider

$${f_4}\left( x \right) - {f_6}\left( x \right)$$

$$=$$ $${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$$

$$= {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]$$

$$= {1 \over 4} - {1 \over 6} = {1 \over {12}}$$
2

### JEE Main 2013 (Offline)

$$ABCD$$ is a trapezium such that $$AB$$ and $$CD$$ are parallel and $$BC \bot CD.$$ If $$\angle ADB = \theta ,\,BC = p$$ and $$CD = q,$$ then AB is equal to:
A
$${{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {p\cos \theta + q\sin \theta }}$$
B
$${{{p^2} + {q^2}\cos \theta } \over {p\cos \theta + q\sin \theta }}$$
C
$${{{p^2} + {q^2}} \over {{p^2}\cos \theta + {q^2}\sin \theta }}$$
D
$${{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {{{\left( {p\cos \theta + q\sin \theta } \right)}^2}}}$$

## Explanation

From Sine Rule

$${{AB} \over {\sin \theta }} = {{\sqrt {{p^2} + {q^2}} } \over {\sin \left( {\pi - \left( {\theta + \alpha } \right)} \right)}}$$

$$AB = {{\sqrt {{p^2} + {q^2}} \sin \theta } \over {\sin \theta \cos \alpha + \cos \theta \sin \alpha }}$$

$$= {{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {q\sin \theta + p\cos \theta }}$$

(As $$\cos \alpha = {q \over {\sqrt {{p^2} + {q^2}} }}$$ and $$\sin \alpha = {p \over {\sqrt {{p^2} + {q^2}} }}$$ )
3

### 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 $$\cot A = {{\cos A} \over {\sin A}}$$ )

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

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

$$= 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$$
4

### 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 $$R = {\pi \over 6}$$

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