If cot2 x - cotx-y - cotx-z where x -4- then cot 2x-

Cot^2 x=cot (x y) . cot(x z) cot^2 x= ( cot x cot y+1/cot y cot x ) ( cot x cot z+1/cot z cot x ) ⇒ cot^2 x (cot y cot x)(cot z cot x)=(cot x cot y+1)(cot x c ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If cot2 x = c...

Question

If $c o t^{2} x = c o t (x - y) . c o t (x - z)$ where $x \neq \pm \frac{π}{4}$ , then $c o t 2 x =$

$\frac{1}{2} (c o t y + c o t z)$

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$\frac{1}{2} (c o t z c o t y)$

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$\frac{1}{2} (c o t z + \frac{1}{c o t^{2}})$

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$(c o t y + \frac{1}{c o t y})$

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Solution

The correct option is A
$\frac{1}{2} (c o t y + c o t z)$

$c o t^{2} x = c o t (x - y) . c o t (x - z) c o t^{2} x = (\frac{c o t x c o t y + 1}{c o t y - c o t x}) (\frac{c o t x c o t z + 1}{c o t z - c o t x}) \Rightarrow c o t^{2} x (c o t y - c o t x) (c o t z - c o t x) = (c o t x c o t y + 1) (c o t x c o t z + 1) \Rightarrow c o t^{3} x (c o t y + c o t z) + c o t x (c o t y + c o t z) + 1 - c o t^{4} x = 0 \Rightarrow c o t x (c o t y + c o t z) (1 + c o t^{2} x) + (1 - c o t^{2} x) (1 + c o t^{2} x) = 0 1 + c o t^{2} x \neq 0 ∴ c o t x (c o t y + c o t z) + (1 - c o t^{2} x) = 0 \Rightarrow \frac{c o t^{2} x - 1}{2 c o t x} = \frac{1}{2} (c o t y + c o t z)$

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Similar questions

Q. Solve the following equations.

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