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

Answer:

Here is the proof of the trigonometric identity

$\frac{\csc x \cos x}{\tan x + \cot x} = \cos^2 x$

Given:

$\frac{\csc x \cos x}{\tan x + \cot x}$

To prove:

$\frac{\csc x \cos x}{\tan x + \cot x} = \cos^2x$

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

$\frac{\csc x \cos x}{\tan x + \cot x} \\ = \frac{\frac{1}{\sin x} \cos x}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}} \\ = \frac{\cos x}{\sin x \cdot \frac{\sin x}{\cos x} + \cos x \cdot \frac{\cos x}{\sin x}} \\ \\ = \frac{\cos x}{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}} \\ \\ = \frac{\cos x}{\frac{1}{\sin x \cos x}} \\ \\ = \cos x \cdot \sin x \cos x \\ \\ = \cos^2 x$

Step 2:Simplify the expression.

$\cos^2 x = \cos^2 x$

Conclusion:Therefore,

$\frac{\csc x \cos x}{\tan x + \cot x} = \cos^2 x.$

Hence Proved:

Answer 2

Answer:

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

$\boxed{\dfrac{ \dfrac{1}{\sin x} \cdot \cos x}{\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}}}=\cos^2 x$

$\boxed{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\sin^2 x }{\sin x + \cos x}+\dfrac{\cos^2 x }{\sin x + \cos x}}}=\cos^2 x$

$\boxed{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\sin^2 x + \cos^2 x}{\sin x + \cos x}}}=\cos^2 x$

$\boxed{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x + \cos x}}}=\cos^2 x$

$\boxed{\dfrac{\cos x}{\sin x} \cdot \sin x \cos x}=\cos^2 x$

$\cos^2 x=\cos^2 x$

Step-by-step explanation:

Given trigonometric identity:

$\dfrac{ \csc x \cos x}{\tan x + \cot x}=\cos^2 x$

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

$\textsf{Rewrite \csc x=\dfrac{1}{\sin x}}, \tan x=\dfrac{\sin x}{\cos x} , and \cot x=\dfrac{\cos x}{\sin x} :}$

$\boxed{\dfrac{ \dfrac{1}{\sin x} \cdot \cos x}{\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x}}}=\cos^2 x$

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

$\boxed{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\sin^2 x }{\sin x + \cos x}+\dfrac{\cos^2 x }{\sin x + \cos x}}}=\cos^2 x$

$\textsf{Apply\;the\;fraction\;rule\;\; \dfrac{a}{b}+\dfrac{c}{b}=\dfrac{a+c}{b} \;to\;the\;denominator}:$

$\boxed{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\sin^2 x + \cos^2 x}{\sin x + \cos x}}}=\cos^2 x$

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

$\boxed{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x + \cos x}}}=\cos^2 x$

$\textsf{Apply\;the\;fraction\;rule\;\; \dfrac{a}{\frac{b}{c}}=a \cdot \dfrac{c}{b} }:$

$\boxed{\dfrac{\cos x}{\sin x} \cdot \sin x \cos x}=\cos^2 x$

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

$\cos^2 x=\cos^2 x$