Question

1. \\(\dfrac{\sec x \csc x - \tan x}{\sec x \csc x}\\)

Explanation:

Step1: Recall trigonometric identities

We know that \(\sec x=\frac{1}{\cos x}\), \(\csc x = \frac{1}{\sin x}\), and \(\tan x=\frac{\sin x}{\cos x}\). Substitute these into the numerator and denominator.
The numerator: \(\sec x\csc x-\tan x=\frac{1}{\cos x}\cdot\frac{1}{\sin x}-\frac{\sin x}{\cos x}=\frac{1 - \sin^{2}x}{\sin x\cos x}\) (by getting a common denominator \(\sin x\cos x\)).
The denominator: \(\sec x\csc x=\frac{1}{\cos x}\cdot\frac{1}{\sin x}=\frac{1}{\sin x\cos x}\).

Step2: Simplify the fraction

Now we have \(\frac{\frac{1 - \sin^{2}x}{\sin x\cos x}}{\frac{1}{\sin x\cos x}}\). When dividing by a fraction, we multiply by its reciprocal, so this becomes \(\frac{1 - \sin^{2}x}{\sin x\cos x}\cdot\frac{\sin x\cos x}{1}\).
The \(\sin x\cos x\) terms cancel out, and we know from the Pythagorean identity that \(1-\sin^{2}x = \cos^{2}x\). So we are left with \(\cos^{2}x\)? Wait, no, wait. Wait, let's re - check. Wait, the numerator after substitution: \(\frac{1}{\cos x\sin x}-\frac{\sin x}{\cos x}=\frac{1-\sin^{2}x}{\cos x\sin x}\)? Wait, no, \(\frac{1}{\cos x\sin x}-\frac{\sin x}{\cos x}=\frac{1 - \sin^{2}x}{\cos x\sin x}\)? Wait, \(\frac{\sin x}{\cos x}=\frac{\sin^{2}x}{\cos x\sin x}\), so \(\frac{1}{\cos x\sin x}-\frac{\sin^{2}x}{\cos x\sin x}=\frac{1 - \sin^{2}x}{\cos x\sin x}=\frac{\cos^{2}x}{\cos x\sin x}=\frac{\cos x}{\sin x}=\cot x\)? Wait, I made a mistake earlier. Let's do it again.

Wait, let's start over. The expression is \(\frac{\sec x\csc x-\tan x}{\sec x\csc x}\). Let's split the fraction: \(\frac{\sec x\csc x}{\sec x\csc x}-\frac{\tan x}{\sec x\csc x}\).

The first term \(\frac{\sec x\csc x}{\sec x\csc x}=1\).

For the second term: \(\frac{\tan x}{\sec x\csc x}\). Substitute \(\tan x=\frac{\sin x}{\cos x}\), \(\sec x=\frac{1}{\cos x}\), \(\csc x=\frac{1}{\sin x}\). So \(\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}\cdot\frac{1}{\sin x}}=\frac{\sin x}{\cos x}\cdot\cos x\sin x=\sin^{2}x\).

So the original expression is \(1-\sin^{2}x\), and by the Pythagorean identity \(1 - \sin^{2}x=\cos^{2}x\)? Wait, no, wait: \(\frac{\sec x\csc x-\tan x}{\sec x\csc x}=1-\frac{\tan x}{\sec x\csc x}\).

\(\frac{\tan x}{\sec x\csc x}=\tan x\cdot\cos x\sin x\) (since \(\frac{1}{\sec x}=\cos x\) and \(\frac{1}{\csc x}=\sin x\)). And \(\tan x=\frac{\sin x}{\cos x}\), so \(\frac{\sin x}{\cos x}\cdot\cos x\sin x=\sin^{2}x\). So \(1 - \sin^{2}x=\cos^{2}x\)? Wait, no, wait, I think I messed up the sign. Wait, the original numerator is \(\sec x\csc x-\tan x\), so when we split the fraction, it's \(\frac{\sec x\csc x}{\sec x\csc x}-\frac{\tan x}{\sec x\csc x}=1 - \frac{\tan x}{\sec x\csc x}\).

Wait, \(\frac{\tan x}{\sec x\csc x}=\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}\cdot\frac{1}{\sin x}}=\frac{\sin x}{\cos x}\times\cos x\sin x=\sin^{2}x\). So \(1-\sin^{2}x = \cos^{2}x\)? But let's check with another approach. Let's convert everything to sine and cosine.

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

So the numerator: \(\frac{1}{\cos x}\cdot\frac{1}{\sin x}-\frac{\sin x}{\cos x}=\frac{1-\sin^{2}x}{\cos x\sin x}=\frac{\cos^{2}x}{\cos x\sin x}=\frac{\cos x}{\sin x}=\cot x\)? Wait, no, \(\frac{1-\sin^{2}x}{\cos x\sin x}=\frac{\cos^{2}x}{\cos x\sin x}=\frac{\cos x}{\sin x}\), and the denominator is \(\frac{1}{\cos x\sin x}\). So \(\frac{\frac{\cos x}{\sin x}}{\frac{1}{\cos x\sin x}}=\frac{\cos x}{\sin x}\times\cos x\sin x=\cos^{2}x\). Wait, now I'm confused. Let's take a value for \(x\), say \(x = \frac{\pi}{4}\).

\(\sec(\frac{\pi}{4})=\sqrt{2}\), \(\csc(\frac{\pi}{4})=\sqrt{2}\), \(\tan(\frac{\pi}{4}) = 1\).

Original expression: \(\frac{\sqrt{2}\times\sqrt{2}-1}{\sqrt{2}\times\sqrt{2}}=\frac{2 - 1}{2}=\frac{1}{2}\).

\(\cos^{2}(\frac{\pi}{4})=(\frac{\sqrt{2}}{2})^{2}=\frac{1}{2}\). Ah, so that works. Wait, but when I did the split, I had a mistake. Let's re - do the split:

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

\(\frac{\tan x}{\sec x\csc x}=\tan x\times\cos x\sin x\) (because \(\frac{1}{\sec x}=\cos x\) and \(\frac{1}{\csc x}=\sin x\)).

\(\tan x=\frac{\sin x}{\cos x}\), so \(\frac{\sin x}{\cos x}\times\cos x\sin x=\sin^{2}x\).

Then \(1-\sin^{2}x=\cos^{2}x\), which matches the value when \(x = \frac{\pi}{4}\).

Wait, but earlier when I thought the numerator simplified to \(\frac{\cos x}{\sin x}\), that was a mistake. The correct simplification of the numerator \(\frac{1}{\cos x\sin x}-\frac{\sin x}{\cos x}\) is \(\frac{1-\sin^{2}x}{\cos x\sin x}=\frac{\cos^{2}x}{\cos x\sin x}=\frac{\cos x}{\sin x}\)? No, wait \(\frac{1}{\cos x\sin x}-\frac{\sin x}{\cos x}=\frac{1 - \sin^{2}x}{\cos x\sin x}\) (because \(\frac{\sin x}{\cos x}=\frac{\sin^{2}x}{\cos x\sin x}\)), so \(1-\sin^{2}x=\cos^{2}x\), so \(\frac{\cos^{2}x}{\cos x\sin x}=\frac{\cos x}{\sin x}\). Then dividing by the denominator \(\frac{1}{\cos x\sin x}\) gives \(\frac{\cos x}{\sin x}\times\cos x\sin x=\cos^{2}x\). Yes, that's correct.

So the simplified form of \(\frac{\sec x\csc x-\tan x}{\sec x\csc x}\) is \(\cos^{2}x\)? Wait, no, wait when \(x=\frac{\pi}{4}\), \(\cos^{2}x=\frac{1}{2}\), and the original expression also equals \(\frac{1}{2}\). Let's try \(x = \frac{\pi}{6}\).

\(\sec(\frac{\pi}{6})=\frac{2}{\sqrt{3}}\), \(\csc(\frac{\pi}{6}) = 2\), \(\tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}\).

Numerator: \(\frac{2}{\sqrt{3}}\times2-\frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}-\frac{1}{\sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\).

Denominator: \(\frac{2}{\sqrt{3}}\times2=\frac{4}{\sqrt{3}}\).

The expression: \(\frac{\sqrt{3}}{\frac{4}{\sqrt{3}}}=\frac{3}{4}\).

\(\cos^{2}(\frac{\pi}{6})=(\frac{\sqrt{3}}{2})^{2}=\frac{3}{4}\). It works. So the simplified form is \(\cos^{2}x\)? Wait, no, \(\cos^{2}(\frac{\pi}{6})=\frac{3}{4}\), which matches. So the correct simplification is \(\cos^{2}x\)? Wait, but let's do it algebraically again.

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

\(\frac{\tan x}{\sec x\csc x}=\tan x\times\cos x\sin x\) (since \(\sec x=\frac{1}{\cos x}\), \(\csc x=\frac{1}{\sin x}\), so \(\frac{1}{\sec x\csc x}=\cos x\sin x\))

\(\tan x=\frac{\sin x}{\cos x}\), so \(\tan x\times\cos x\sin x=\frac{\sin x}{\cos x}\times\cos x\sin x=\sin^{2}x\)

Thus, \(1-\sin^{2}x=\cos^{2}x\) (by the Pythagorean identity \(\sin^{2}x+\cos^{2}x = 1\))

Answer:

\(\cos^{2}x\)