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sin 2x (tan x + cot x) = ?...

`sin 2x (tan x + cot x) = ?`

A

`sin x cos x`

B

`cosec x sec x`

C

`2tan(2x)`

D

`2`

Text Solution

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The correct Answer is:
To solve the equation \( \sin 2x (\tan x + \cot x) \), we will follow these steps: ### Step 1: Rewrite \( \sin 2x \) Using the double angle identity for sine, we have: \[ \sin 2x = 2 \sin x \cos x \] ### Step 2: Rewrite \( \tan x + \cot x \) We can express \( \tan x \) and \( \cot x \) in terms of sine and cosine: \[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \] Thus, \[ \tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \] ### Step 3: Find a common denominator To combine the fractions, we find a common denominator, which is \( \sin x \cos x \): \[ \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \] ### Step 4: Use the Pythagorean identity From the Pythagorean identity, we know: \[ \sin^2 x + \cos^2 x = 1 \] Thus, \[ \tan x + \cot x = \frac{1}{\sin x \cos x} \] ### Step 5: Substitute back into the equation Now we substitute back into our original equation: \[ \sin 2x (\tan x + \cot x) = (2 \sin x \cos x) \left( \frac{1}{\sin x \cos x} \right) \] ### Step 6: Simplify the expression The \( \sin x \cos x \) terms cancel out: \[ \sin 2x (\tan x + \cot x) = 2 \] ### Conclusion Thus, we have shown that: \[ \sin 2x (\tan x + \cot x) = 2 \]

To solve the equation \( \sin 2x (\tan x + \cot x) \), we will follow these steps: ### Step 1: Rewrite \( \sin 2x \) Using the double angle identity for sine, we have: \[ \sin 2x = 2 \sin x \cos x \] ...
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Knowledge Check

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    `alpha ^(2) + 2 alpha + 1=0`
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    B
    `(1)/(20)tan^(-1)((1)/(9sqrt(3)))`
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    `(1)/(10)((pi)/(4)-tan^(-1)((1)/(9sqrt(3))))`
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