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 .


The diagram makes it easier. It is consistent with your given conditions with $\frac ba = \tan \theta$. Note that $c = \sqrt{a^2 +b^2}$ (Pythagoras).
From the diagram, $\sin \theta = \frac bc, \cos \theta = \frac ac$.
So $b\sin \theta + a\cos \theta = \frac {b^2}c + \frac {a^2} c = \frac {c^2}c = c = \sqrt{a^2 + b^2}$ (QED)
