<s>[INST] How can I derive $~\text{opposite}\cdot\sin^{}\left(\theta_{}\right)+\text{adjacent}\cdot\cos^{}\left(\theta_{}\right)=\tan^{}\left(\theta_{}\right)$ Given the below equation .
$$  b \cos^{}\left(\theta_{} \right) = a \cdot \sin^{}\left(\theta_{} \right)  $$
I have to derive the below equation .
$$  b \sin^{}\left(\theta_{} \right) + a \cdot \cos^{}\left(\theta_{} \right) = \sqrt{ a ^2 + b^2 }  $$
My tries are as below .
$$  \frac{  b   }{  a  } = \frac{  \sin^{}\left(\theta_{} \right)   }{  \cos^{}\left(\theta_{} \right)   }  $$
$$  \frac{  b   }{  a  } = \tan^{}\left( \theta_{}  \right)   $$
$$  \text{adjacent}= a   $$
$$  \text{opposite}= b   $$
$$  \text{hypotenuse}=\sqrt{ a^2+ b ^2 }   $$
$$  b \sin^{}\left(\theta_{} \right) + a \cdot \cos^{}\left(\theta_{} \right)  $$
$$ = b \left( \frac{  b \cos^{}\left(\theta_{} \right)   }{  a  }  \right) +a \left( \frac{  a \sin^{}\left(\theta_{} \right)   }{  b   }  \right)  $$
$$ = \frac{  b^2\cos^{}\left(\theta_{} \right)   }{  a  } + \frac{  a ^2 \sin^{}\left(\theta_{} \right)   }{  b   }  $$
I've been got stucked from here .
 [/INST] Of the last line .
$$  \frac{  b^2\cos^{}\left(\theta_{} \right)   }{  a  } + \frac{  a ^2 \sin^{}\left(\theta_{} \right)   }{  b   }  $$
Just plug the below ones to the above formula and you can get the RHS of the problem .
$$  \sin^{}\left(\theta_{} \right) = \frac{  b  }{ \sqrt{ a^2+b^2 }    }  $$
$$  \cos^{}\left(\theta_{} \right) = \frac{  a  }{ \sqrt{ a^2+b^2 }    }  $$
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