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\frac{1+\tan \theta}{1+\cot \theta} = \tan \theta
Question
\frac{1+\tan\left(\theta \right)}{1+\cot\left(\theta \right)}=\tan\left(\theta \right)
Uh oh!
Solve the equation
\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}
Alternative Form
\theta \neq \left\{ \begin{array}{l}90^{\circ} k\\135^{\circ}+180^{\circ} k\end{array}\right.,k \in \mathbb{Z}
Evaluate
\frac{1+\tan\left(\theta \right)}{1+\cot\left(\theta \right)}=\tan\left(\theta \right)
Find the domain
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Evaluate
\left\{ \begin{array}{l}\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\\\theta \neq k\pi ,k \in \mathbb{Z}\\1+\cot\left(\theta \right)\neq 0\end{array}\right.
Calculate
\left\{ \begin{array}{l}\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\\\theta \neq k\pi ,k \in \mathbb{Z}\\\theta \neq \frac{3\pi }{4}+k\pi ,k \in \mathbb{Z}\end{array}\right.
Find the intersection
\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}
\frac{1+\tan\left(\theta \right)}{1+\cot\left(\theta \right)}=\tan\left(\theta \right),\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}
Rewrite the expression
\frac{1+\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}}{1+\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}}=\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}
Calculate
\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}=\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}
\text{The statement is true for any value of }\theta
\theta \in \mathbb{R}
Check if the solution is in the defined range
\theta \in \mathbb{R},\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}
Solution
\theta \neq \left\{ \begin{array}{l}\frac{k\pi }{2}\\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}
Alternative Form
\theta \neq \left\{ \begin{array}{l}90^{\circ} k\\135^{\circ}+180^{\circ} k\end{array}\right.,k \in \mathbb{Z}
Verify the identity
\textrm{true}
Evaluate
\frac{1+\tan\left(\theta \right)}{1+\cot\left(\theta \right)}=\tan\left(\theta \right)
Start working on the left-hand side
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Evaluate
\frac{1+\tan\left(\theta \right)}{1+\cot\left(\theta \right)}
Transform the expression
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Evaluate
1+\tan\left(\theta \right)
\text{Use }\tan t = \frac{\sin t}{\cos t}\text{ to transform the expression}
1+\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}
Reduce fractions to a common denominator
\frac{\cos\left(\theta \right)}{\cos\left(\theta \right)}+\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}
Write all numerators above the common denominator
\frac{\cos\left(\theta \right)+\sin\left(\theta \right)}{\cos\left(\theta \right)}
\frac{\frac{\cos\left(\theta \right)+\sin\left(\theta \right)}{\cos\left(\theta \right)}}{1+\cot\left(\theta \right)}
Transform the expression
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Evaluate
1+\cot\left(\theta \right)
\text{Use }\cot t = \frac{\cos t}{\sin t}\text{ to transform the expression}
1+\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}
Reduce fractions to a common denominator
\frac{\sin\left(\theta \right)}{\sin\left(\theta \right)}+\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)}
Write all numerators above the common denominator
\frac{\sin\left(\theta \right)+\cos\left(\theta \right)}{\sin\left(\theta \right)}
\frac{\frac{\cos\left(\theta \right)+\sin\left(\theta \right)}{\cos\left(\theta \right)}}{\frac{\sin\left(\theta \right)+\cos\left(\theta \right)}{\sin\left(\theta \right)}}
Multiply by the reciprocal
\frac{\cos\left(\theta \right)+\sin\left(\theta \right)}{\cos\left(\theta \right)}\times \frac{\sin\left(\theta \right)}{\sin\left(\theta \right)+\cos\left(\theta \right)}
\text{Cancel out the common factor }\cos\left(\theta \right)+\sin\left(\theta \right)
\frac{1}{\cos\left(\theta \right)}\times \sin\left(\theta \right)
Multiply the terms
\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}
\text{Use }\frac{\sin(t)}{\cos(t)}=\tan(t)\text{ to transform the expression}
\tan\left(\theta \right)
\tan\left(\theta \right)=\tan\left(\theta \right)
Solution
\textrm{true}
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