How do you prove $(sin x - tan x) (cos x - cot x) = (sin x - 1) (cos x - 1)$ $(sinx - tanx)(cosx - cotx) = (sinx -1)(cosx -1)$ ?

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How do you prove $(sin x - tan x) (cos x - cot x) = (sin x - 1) (cos x - 1)$ $(sinx - tanx)(cosx - cotx) = (sinx -1)(cosx -1)$ ?

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Answer 1 · 2018-04-03T08:13:34

$L . H . S = (sin x - tan x) (cos x - cot x)$ $L.H.S = (sinx−tanx)(cosx−cotx)$

$\Rightarrow (sin x - \frac{sin x}{cos x}) (cos x - \frac{cos x}{sin x})$ $=> (sinx−sinx/cosx)(cosx−cosx/sinx)$

$\Rightarrow (\frac{sin x cos x - sin x}{cos x}) (\frac{cos x sin x - cos x}{sin x})$ $=> ((sinxcosx−sinx)/cosx)((cosxsinx−cosx)/sinx)$

$\Rightarrow \frac{1}{cos x sin x} (sin x cos x - sin x) (cos x sin x - cos x)$ $=> 1/(cosxsinx)(sinxcosx−sinx)(cosxsinx−cosx)$

$\Rightarrow \frac{1}{cos x sin x} ({sin}^{2} x {cos}^{2} x - {cos}^{2} x sin x - {sin}^{2} x cos x + sin x cos x)$ $=> 1/(cosxsinx)(sin^2xcos^2x-cos^2xsinx-sin^2xcosx+sinxcosx)$

$\Rightarrow (sin x cos x - cos x - sin x + 1)$ $=> (sinxcosx-cosx-sinx+1)$

$\Rightarrow cos x (sin x - 1) - 1 (sin x - 1)$ $=> cosx(sinx-1) -1(sinx-1)$

$\Rightarrow (cos x - 1) (sin x - 1)$ $=>( cosx-1)(sinx-1)$

Answer 2 · 2018-04-03T10:54:51

$L . H . S = (sin x - tan x) (cos x - cot x)$ $L.H.S = (sinx−tanx)(cosx−cotx)$

$= tan x (\frac{sin x}{tan x} - \frac{tan x}{tan x}) \cdot cot x (\frac{cos x}{cot x} - \frac{cot x}{cot x})$ $=tanx (sinx/tanx−tanx/tanx)*cotx(cosx/cotx−cotx/cotx)$

$= tan x \cdot cot x (\frac{sin x}{\frac{sin x}{cos x}} - 1) (\frac{cos x}{\frac{cos x}{sin x}} - 1)$ $=tanx*cotx (sinx/(sinx/cosx)−1)(cosx/(cosx/sinx)−1)$

$= (cos x - 1) (sin x - 1) = R H S$ $=(cosx-1)(sinx-1)=RHS$

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How do you prove $(sin x - tan x) (cos x - cot x) = (sin x - 1) (cos x - 1)$ $(sinx - tanx)(cosx - cotx) = (sinx -1)(cosx -1)$ ?

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