## Trigonometry (11th Edition) Clone

$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ The equation is an identity, as verified below.
$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ 1) We start from the left side. $$X=\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)$$ - Recall the sum identity for tangent, which states $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ Replace the identity into $X$ with $A=\frac{x}{2}$ and $B=\frac{\pi}{4}$, we have $$X=\frac{\tan\frac{x}{2}+\tan\frac{\pi}{4}}{1-\tan\frac{x}{2}\tan\frac{\pi}{4}}$$ $$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}\times1}$$ $$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}}$$ - Now use the half-angle identity for tangent: $$\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$$ Thus, $$X=\frac{\frac{1-\cos x}{\sin x}+1}{1-\frac{1-\cos x}{\sin x}}$$ $$X=\frac{\frac{1-\cos x+\sin x}{\sin x}}{\frac{\sin x-1+\cos x}{\sin x}}$$ $$X=\frac{1-\cos x+\sin x}{\sin x-1+\cos x}$$ $$X=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$ We stop here as there seems no potential in simplifying anymore. 2) Then we look at the right side. $$Y=\sec x+\tan x$$ - Apply the following identities: $$\sec x=\frac{1}{\cos x}\hspace{2cm}\tan x=\frac{\sin x}{\cos x}$$ $$Y=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$$ $$Y=\frac{1+\sin x}{\cos x}$$ So the job here is to prove that $\frac{1+\sin x}{\cos x}=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$, which seems so different. So in a seemingly dead-end case like this, we would try multiplying both the numerator and denominator of $Y$ with $\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$. $$Y=\frac{1+\sin x}{\cos x}\times\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$$ $$Y=\frac{\sin x+\cos x-1+\sin^2x+\sin x\cos x-\sin x}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\sin^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$ - Now remember that $\sin^2x=1-\cos^2x$, which can be used to rewrite $\sin^2x$. $$Y=\frac{1-\cos^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{-\cos^2x+\sin x\cos x+\cos x}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\cos x(\sin x-\cos x+1)}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$ So after a rather painstaking process of transformations, we show that $X=Y$. Therefore, $$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ The equation is therefore an identity.
The text passage provides a detailed proof of a trigonometric identity, showing step-by-step how the left side of the equation is transformed to match the right side.
no_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## Trigonometry (11th Edition) Clone

$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ The equation is an identity, as verified below.
$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ 1) We start from the left side. $$X=\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)$$ - Recall the sum identity for tangent, which states $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ Replace the identity into $X$ with $A=\frac{x}{2}$ and $B=\frac{\pi}{4}$, we have $$X=\frac{\tan\frac{x}{2}+\tan\frac{\pi}{4}}{1-\tan\frac{x}{2}\tan\frac{\pi}{4}}$$ $$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}\times1}$$ $$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}}$$ - Now use the half-angle identity for tangent: $$\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$$ Thus, $$X=\frac{\frac{1-\cos x}{\sin x}+1}{1-\frac{1-\cos x}{\sin x}}$$ $$X=\frac{\frac{1-\cos x+\sin x}{\sin x}}{\frac{\sin x-1+\cos x}{\sin x}}$$ $$X=\frac{1-\cos x+\sin x}{\sin x-1+\cos x}$$ $$X=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$ We stop here as there seems no potential in simplifying anymore. 2) Then we look at the right side. $$Y=\sec x+\tan x$$ - Apply the following identities: $$\sec x=\frac{1}{\cos x}\hspace{2cm}\tan x=\frac{\sin x}{\cos x}$$ $$Y=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$$ $$Y=\frac{1+\sin x}{\cos x}$$ So the job here is to prove that $\frac{1+\sin x}{\cos x}=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$, which seems so different. So in a seemingly dead-end case like this, we would try multiplying both the numerator and denominator of $Y$ with $\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$. $$Y=\frac{1+\sin x}{\cos x}\times\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$$ $$Y=\frac{\sin x+\cos x-1+\sin^2x+\sin x\cos x-\sin x}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\sin^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$ - Now remember that $\sin^2x=1-\cos^2x$, which can be used to rewrite $\sin^2x$. $$Y=\frac{1-\cos^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{-\cos^2x+\sin x\cos x+\cos x}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\cos x(\sin x-\cos x+1)}{\cos x(\sin x+\cos x-1)}$$ $$Y=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$ So after a rather painstaking process of transformations, we show that $X=Y$. Therefore, $$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$ The equation is therefore an identity.
user
The text passage provides a detailed proof of a trigonometric identity, showing step-by-step how the left side of the equation is transformed to match the right side.

TYPE: NO_QUESTION