How do you prove $(cos x - 1) (tan x - 1) = 0$ $(cosx-1)(tanx-1)=0$ ?

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How do you prove $(cos x - 1) (tan x - 1) = 0$ $(cosx-1)(tanx-1)=0$ ?

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Answer 1 · 2018-04-12T11:09:52

Please check the Question.

Explanation:

Here,
$(cos x - 1) (tan x - 1) = 0$ $(cosx-1)(tanx-1)=0$
If we take , $x = 30^{\circ},$ $x=30^circ ,$ then
$L H S = ({cos 30}^{\circ} - 1) ({tan 30}^{\circ} - 1)$ $LHS=(cos30^circ-1)(tan30^circ-1)$
$= (\frac{\sqrt{3}}{2} - 1) (\frac{1}{\sqrt{3}} - 1)$ $=(sqrt3/2-1)(1/sqrt3-1)$
$\approx (0.87 - 1) (0.58 - 1)$ $~~(0.87-1)(0.58-1)$
$\approx (- 0.13) (- 0.42)$ $~~(-0.13)(-0.42)$
$\approx 0.055 \neq 0$ $~~0.055!=0$
$L H S \neq R H S$ $LHS!=RHS$

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How do you prove $(cos x - 1) (tan x - 1) = 0$ $(cosx-1)(tanx-1)=0$ ?

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How do you prove (cosx-1)(tanx-1)=0(cosx−1)(tanx−1)=0(cosx-1)(tanx-1)=0?

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How do you prove $(cos x - 1) (tan x - 1) = 0$ $(cosx-1)(tanx-1)=0$ ?