How do you verify $\frac{csc (- x)}{sec (- x)} = - cot x$ $csc (-x) / sec ( - x) =- cot x$ ?

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Answer 1 · 2016-07-13T05:37:20

For this problem, we can make use of some trigonometric identities such as these:

$sin (- x) = - sin (x)$ $sin(-x) = -sin(x)$
$cos (- x) = cos (x)$ $cos(-x) = cos(x)$
$tan (- x) = - tan (x)$ $tan(-x) = -tan(x)$

$csc (x) = \frac{1}{sin (x)}$ $csc(x) = (1)/(sin(x))$

$sec (x) = \frac{1}{cos (x)}$ $sec(x) = (1)/(cos(x))$

$cot (x) = \frac{1}{tan (x)} = \frac{cos (x)}{sin (x)}$ $cot(x) = (1)/(tan(x)) = (cos(x))/(sin(x))$

We can now try to rewrite our original equation, starting from the left-hand side, which gives us

Left-Hand Side:

$(csc(-x))/(sec(-x)) = ((1)/(-sin x))/ ((1)/(cos x)) = ((1)/(-sinx))/ ((1)/(cancel(cos x))) *(cos x/1)/(cancel(cos x)/1) =- (cosx)/(sinx$

Right-Hand Side:

$-cot(x) = (1)/(-tan x) = - cos x / sin x$

Since the left-hand side equals the right-hand side, we have verified that

$csc(-x)/(sec(-x)) = -cot(x)$

How do you verify csc (-x) / sec ( - x) =- cot xcsc(−x)sec(−x)=−cotxcsc (-x) / sec ( - x) =- cot x?