Question

Pls help

Prove the trigonometric identity
csc x cos x/tan x+cot x =cos^2 x
Drag the expressions into each box to complete the proof.
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Answer 1

Here is the proof of the trigonometric identity

`(\csc x \cos x)/(\tan x + \cot x) = \cos^2 x`

Given:

`(\csc x \cos x)/(\tan x + \cot x)`

To prove:

`(\csc x \cos x)/(\tan x + \cot x) = \cos^2x`

Step 1: Rewrite all trigonometric functions in terms of sine and cosine.

`(\csc x \cos x)/(\tan x + \cot x) \\ = ((1)/(\sin x) \cos x)/((\sin x)/(\cos x) + (\cos x)/(\sin x)) \\ = (\cos x)/(\sin x \cdot (\sin x)/(\cos x) + \cos x \cdot (\cos x)/(\sin x)) \\ \\ = (\cos x)/((\sin^2 x + \cos^2 x)/(\sin x \cos x)) \\ \\ = (\cos x)/((1)/(\sin x \cos x)) \\ \\ = \cos x \cdot \sin x \cos x \\ \\ = \cos^2 x`

Step 2:Simplify the expression.

`\cos^2 x = \cos^2 x`

Conclusion:Therefore,

`(\csc x \cos x)/(\tan x + \cot x) = \cos^2 x.`

Hence Proved:

Answer 2

`( \csc x \cos x)/(\tan x + \cot x)=\cos^2 x`
`\textsf{Rewrite $\csc x=(1)/(\sin x)}, \tan x=(\sin x)/(\cos x)$, and $\cot x=(\cos x)/(\sin x)$:}`

`\boxed{( (1)/(\sin x) \cdot \cos x)/((\sin x)/(\cos x) + (\cos x)/(\sin x))}=\cos^2 x`

`\boxed{((\cos x)/(\sin x))/((\sin^2 x )/(\sin x + \cos x)+(\cos^2 x )/(\sin x + \cos x))}=\cos^2 x`

`\boxed{((\cos x)/(\sin x))/((\sin^2 x + \cos^2 x)/(\sin x + \cos x))}=\cos^2 x`

`\boxed{((\cos x)/(\sin x))/((1)/(\sin x + \cos x))}=\cos^2 x`

`\boxed{(\cos x)/(\sin x) \cdot \sin x \cos x}=\cos^2 x`

`\cos^2 x=\cos^2 x`

Explanation:

Given trigonometric identity:

`( \csc x \cos x)/(\tan x + \cot x)=\cos^2 x`

`\textsf{Use the identities\;\;$\csc x=(1)/(\sin x)$\;,\;$\tan x = (\sin x)/(\cos x)$\;\;and\;\;$\cot x=(\cos x)/(\sin x)$}:`

`\textsf{Rewrite $\csc x=(1)/(\sin x)}, \tan x=(\sin x)/(\cos x)$, and $\cot x=(\cos x)/(\sin x)$:}`

`\boxed{( (1)/(\sin x) \cdot \cos x)/((\sin x)/(\cos x) + (\cos x)/(\sin x))}=\cos^2 x`

Simplify the numerator and make the fractions in the denominator like fractions:

`\boxed{((\cos x)/(\sin x))/((\sin^2 x )/(\sin x + \cos x)+(\cos^2 x )/(\sin x + \cos x))}=\cos^2 x`

`\textsf{Apply\;the\;fraction\;rule\;\;$(a)/(b)+(c)/(b)=(a+c)/(b)$\;to\;the\;denominator}:`

`\boxed{((\cos x)/(\sin x))/((\sin^2 x + \cos^2 x)/(\sin x + \cos x))}=\cos^2 x`

`\textsf{Use\;the\;identity\;\;$\sin^2x+\cos^2x=1$}:`

`\boxed{((\cos x)/(\sin x))/((1)/(\sin x + \cos x))}=\cos^2 x`

`\textsf{Apply\;the\;fraction\;rule\;\;$(a)/((b)/(c))=a \cdot (c)/(b)$}:`

`\boxed{(\cos x)/(\sin x) \cdot \sin x \cos x}=\cos^2 x`

Cancel the common factor sin x, and apply the exponent rule aa = a²:

`\cos^2 x=\cos^2 x`