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1

WB JEE 2010

MCQ (Single Correct Answer)

Value of $${\tan ^{ - 1}}\left( {{{\sin 2 - 1} \over {\cos 2}}} \right)$$ is

A
$${\pi \over 2} - 1$$
B
$$1 - {\pi \over 4}$$
C
$$2 - {\pi \over 2}$$
D
$${\pi \over 4} - 1$$

Explanation

$$ - {\tan ^{ - 1}}\left( {{{1 - \sin 2} \over {\cos 2}}} \right)$$ [$$\because$$ $${\tan ^{ - 1}}( - x) = - {\tan ^{ - 1}}x$$]

$$ = - {\tan ^{ - 1}}\left\{ {{{{{(\cos 1 - sin1)}^2}} \over {{{\cos }^2}1 - {{\sin }^2}1}}} \right\} = - {\tan ^{ - 1}}\left( {{{\cos 1 - \sin 1} \over {\cos 1 + \sin 1}}} \right)$$

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

2

WB JEE 2009

MCQ (Single Correct Answer)

$$\tan \left[ {{\pi \over 4} + {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right] + \tan \left[ {{\pi \over 4} - {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right]$$ is equal to

A
$${{2a} \over b}$$
B
$${{2b} \over a}$$
C
$${a \over b}$$
D
$${b \over a}$$

Explanation

$$\tan \left[ {{\pi \over 4} + {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right] + \tan \left[ {{\pi \over 4} - {1 \over 2}{{\cos }^{ - 1}}\left( {{a \over b}} \right)} \right]$$

$$ = \tan \left( {{\pi \over 4} + x} \right) + \tan \left( {{\pi \over 4} - x} \right)$$

Let $${1 \over 2}{\cos ^{ - 1}}{a \over b} = x = {{1 + \tan x} \over {1 - \tan x}} + {{1 - \tan x} \over {1 + \tan x}} = {{2(1 + {{\tan }^2}x)} \over {1 - {{\tan }^2}x}}$$

$$ = {2 \over {\cos 2x}} = {2 \over {a/b}} = {{2b} \over a}$$

3

WB JEE 2008

MCQ (Single Correct Answer)

The value of $$\tan \alpha + 2\tan (2\alpha ) + 4\tan (4\alpha ) + ... + {2^{n - 1}}\tan ({2^{n - 1}}\alpha ) + {2^n}\cot ({2^n}\alpha )$$ is

A
$$\cot ({2^n}\alpha )$$
B
$${2^n}\tan ({2^n}\alpha )$$
C
0
D
$$\cot \alpha $$

Explanation

$$\cot \alpha - \tan \alpha = 2\cot 2\alpha $$

$$2(\cot 2\alpha - \tan 2\alpha ) = {2^2}\cot {2^2}\alpha $$

$${2^2}(\cot {2^2}\alpha - \tan {2^2}\alpha ) = {2^3}\cot {2^3}\alpha $$

$$ \vdots $$

By adding all identities we get

$$\cot \alpha - \tan \alpha - 2\tan 2\alpha - {2^2}{\tan ^2}{2^2}\alpha {...2^{n - 1}}\tan {2^{n - 1}}\alpha = {2^n}\cot {2^n}\alpha $$

$$ \Rightarrow \tan \alpha + 2\tan (2\alpha ) + 4\tan (4\alpha ) + ... + {2^{n - 1}}\tan ({2^{n - 1}}\alpha ) + {2^n}\cot {2^n}\alpha = \cot \alpha $$

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