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4 years ago

Given: ABCD is a trapezoid. BA = CD Prove: BD = CA

Mathematics

1 answer:

lions [1.4K]4 years ago

4 0

Answer:

Step-by-step explanation:

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4 years ago

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3 years ago

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3 years ago

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Write the recurring decimal in its simplest form

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4 years ago

Verify the identity. cotangent of x divided by quantity one plus cosecant of x equals quantity cosecant of x minus one divided b

Elenna [48]

Answer:

$\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}$

Step-by-step explanation:

We want to verify the identity:

$\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}$

Let us take the LHS and simplify to get the LHS.

Express everything in terms of the cosine and sine function.

$\frac{\cot x}{1+\csc x}=\frac{\frac{\cos x}{\sin x} }{1+\frac{1}{\sin x} }$

Collect LCM

$\frac{\cot x}{1+\csc x}=\frac{\frac{\cos x}{\sin x} }{\frac{\sin x+1}{\sin x} }$

We simplify the RHS to get:

$\frac{\cot x}{1+\csc x}=\frac{\cos x}{\sin x+1}$

We rationalize to get:

$\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{(\sin x+1)*(\sin x-1)}$

We expand to get:

$\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{\sin^2 x-1}$

Factor negative one in the denominator:

$\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{-(1-\sin^2 x)}$

Apply the Pythagoras Property to get:

$\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{-\cos^2 x}$

Simplify to get:

$\frac{\cot x}{1+\csc x}=\frac{-(\sin x-1)}{\cos x}$

$\frac{\cot x}{1+\csc x}=\frac{1-\sin x}{\cos x}$

Divide both the numerator and denominator by sin x

$\frac{\cot x}{1+\csc x}=\frac{\frac{1}{\sin x}-\frac{\sin x}{\sin x}}{\frac{\cos x}{\sin x}}$

This finally gives:

$\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}$

8 0

3 years ago