The correct option is A sin 2x = 4/( 2n + 1)π Given #160; tan x=tan x tan x=tan(π 2 tan x) x=nπ +π 2 tan x tan x+ x=nπ+π 2 sin xc ...

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Linear Differential Equations of First Order

If tan x = ...

Question

If $tan (cot x) = cot (tan x),$ then

sin 2 x = \frac{4}{(2 n + 1) π}

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sin 2 x = \frac{4}{π}

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sin 2 x = \frac{1}{(2 n + 1)}

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sin 2 x = n π

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Solution

The correct option is A $sin 2 x = \frac{4}{(2 n + 1) π}$
Given

$tan cot x = cot tan x$

$⟹ tan cot x = tan (\frac{π}{2} - tan x)$

$⟹ cot x = n π + \frac{π}{2} - tan x$

$⟹ tan x + cot x = n π + \frac{π}{2}$

$⟹ \frac{sin x}{cos x} + \frac{cos x}{sin x} = n π + \frac{π}{2}$

$⟹ \frac{{sin}^{2} x + {cos}^{2} x}{sin x cos x} = n π + \frac{π}{2}$

$⟹ 1 = (2 n + 1) \frac{π}{2} sin x cos x$

$⟹ sin x cos x = \frac{2}{(2 n + 1) π}$

$⟹ sin 2 x = \frac{4}{(2 n + 1) π}$

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Similar questions

\frac{1 + sin 2 x}{1 - sin 2 x} = {cot}^{2} (a + x) \forall x \in R \sim (n π + \frac{π}{4}), n \in N,

(1 - tan x) (1 + sin 2 x) = 1 + tan x

n π - \frac{π}{k}, n π

n \in Z

{cos}^{2} 2 x - {sin}^{2} x = 0

\frac{1 + sin 2 x}{1 - sin 2 x} = {cot}^{2} (a + x)

\forall x \in R - (n π + \frac{π}{4}),

n \in N

a

f (x) = s i n^{2} x + s i n^{2} (x + \frac{π}{3}) + c o s x c o s (x + \frac{π}{3}) a n d g (\frac{5}{4}) = 1

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