Question

precalc
verifying trig identities day 2
name:
period: 3

1. \\(\frac{\sin x}{\tan x} = \cos x\\)

\\(\frac{\sin}{\frac{\sin}{\cos}} = \cos\\)

1. \\(\frac{\cos x \sec x}{\tan x} = \cot x\\)

\\(\frac{\cos (\frac{1}{\cos})}{\frac{\sin}{\cos}} = \frac{1}{\frac{\sin}{\cos}} = \frac{\cos}{\sin}\\)

1. \\(\frac{\tan x}{\csc x} = \sec x - \cos x\\)

\\(\frac{\frac{\sin}{\cos}}{\frac{1}{\sin}} = \frac{1}{\cos} - \frac{\cos}{1}\\)

1. \\(\csc x = \sin x + \cos x \cot x\\)

\\(\frac{1}{\sin} = \sin + \cos (\frac{1}{\frac{\sin}{\cos}})\\)

1. \\(\frac{\sec x - \cos x}{\sec x} = \sin^2 x\\)

\\(\frac{\frac{1}{\cos} - \frac{\cos}{1}}{\frac{1}{\cos}} = \sin^2 x\\)

Explanation:

Let's verify each trigonometric identity step by step. We'll start with problem 1:

Problem 1: $\boldsymbol{\frac{\sin x}{\tan x} = \cos x}$

Step 1: Recall the definition of $\tan x$

We know that $\tan x = \frac{\sin x}{\cos x}$. Substitute this into the left - hand side (LHS) of the equation.
The LHS is $\frac{\sin x}{\tan x}$, so substituting $\tan x=\frac{\sin x}{\cos x}$, we get $\frac{\sin x}{\frac{\sin x}{\cos x}}$.

Step 2: Simplify the complex fraction

When we divide by a fraction, we multiply by its reciprocal. So $\frac{\sin x}{\frac{\sin x}{\cos x}}=\sin x\times\frac{\cos x}{\sin x}$.
The $\sin x$ terms in the numerator and denominator cancel out (assuming $\sin x
eq0$, which is a valid assumption for the domain where the original functions are defined), and we are left with $\cos x$, which is equal to the right - hand side (RHS) of the equation.

Problem 2: $\boldsymbol{\frac{\cos x\sec x}{\tan x}=\cot x}$

Step 1: Recall the definitions of $\sec x$ and $\tan x$

We know that $\sec x=\frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$, and $\cot x=\frac{\cos x}{\sin x}$. First, simplify the numerator $\cos x\sec x$.
Substitute $\sec x=\frac{1}{\cos x}$ into $\cos x\sec x$: $\cos x\times\frac{1}{\cos x} = 1$. So the LHS becomes $\frac{1}{\tan x}$.

Step 2: Substitute the definition of $\tan x$ and simplify

Since $\tan x=\frac{\sin x}{\cos x}$, then $\frac{1}{\tan x}=\frac{\cos x}{\sin x}$, and $\frac{\cos x}{\sin x}=\cot x$, which is equal to the RHS.

Problem 3: $\boldsymbol{\frac{\tan x}{\csc x}=\sec x - \cos x}$

Step 1: Recall the definitions of $\tan x$, $\csc x$, and $\sec x$

We know that $\tan x=\frac{\sin x}{\cos x}$, $\csc x=\frac{1}{\sin x}$, and $\sec x=\frac{1}{\cos x}$. First, simplify the LHS.
The LHS is $\frac{\tan x}{\csc x}$, substituting the definitions we get $\frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x}}$.

Step 2: Simplify the complex fraction

When dividing by a fraction, multiply by its reciprocal: $\frac{\sin x}{\cos x}\times\sin x=\frac{\sin^{2}x}{\cos x}$.

Step 3: Use the Pythagorean identity $\sin^{2}x=1 - \cos^{2}x$

Rewrite $\frac{\sin^{2}x}{\cos x}$ as $\frac{1-\cos^{2}x}{\cos x}$. Then, split the fraction: $\frac{1}{\cos x}-\frac{\cos^{2}x}{\cos x}$.
Simplify $\frac{\cos^{2}x}{\cos x}=\cos x$, so we have $\frac{1}{\cos x}-\cos x$, and $\frac{1}{\cos x}=\sec x$, so $\sec x - \cos x$, which is equal to the RHS.

Problem 4: $\boldsymbol{\csc x=\sin x+\cos x\cot x}$

Answer:

Step 1: Recall the definition of $\sec x$

We know that $\sec x=\frac{1}{\cos x}$. First, simplify the numerator of the LHS: $\sec x-\cos x=\frac{1}{\cos x}-\cos x$.

Step 2: Get a common denominator for the numerator

The common denominator is $\cos x$, so $\frac{1}{\cos x}-\cos x=\frac{1 - \cos^{2}x}{\cos x}$.

Step 3: Simplify the fraction

The LHS is $\frac{\frac{1 - \cos^{2}x}{\cos x}}{\frac{1}{\cos x}}$. When dividing by a fraction, multiply by its reciprocal: $\frac{1 - \cos^{2}x}{\cos x}\times\cos x$.
The $\cos x$ terms cancel out, and we are left with $1-\cos^{2}x$.

Step 4: Use the Pythagorean identity $\sin^{2}x=1 - \cos^{2}x$

Since $1-\cos^{2}x=\sin^{2}x$, the LHS is equal to the RHS.

All the trigonometric identities are verified.