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\frac{1+\tan \theta}{1+\cot \theta} = \tan \theta

Algebra Calculus Trigonometry Matrix

Question

\frac{1+\tan\left(\theta \right)}{1+\cot\left(\theta \right)}=\tan\left(\theta \right)

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{\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)}}=\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}

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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)}

Add or subtract the terms

\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)}

Add or subtract the terms

\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)}

Reduce the fraction

\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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