<s>[INST] If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:     
$$ 40\cosθ+9\sinθ=41 $$ 
$$ 9\sinθ=41-40\cos\theta   $$   
Squaring both the sides:   
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$ 
$$81-81 \cos^2\theta=  1681+1600\cos^2\theta-3280 \cos\theta$$
$$1681\cos^2\theta-3280\cos\theta+1600=0$$   
Solving the quadratic equation gives  $\cos\theta=\dfrac{40}{41}$. It is not easy to solve the quadratic equation without calculator so there must be some other method, if yes then please explain.   
P.S: I've found the other method so I am self-answering the question.
 [/INST] Since $41^2 - 40^2 = (41-40)(41+40) = 81 = 9^2$, there exists an angle $\phi$ satisfying $$\sin \phi = \frac{40}{41}, \quad \cos \phi = \frac{9}{41},$$ hence $$\begin{align*} 1 &= \frac{9}{41} \sin \theta + \frac{40}{41} \cos \theta \\ &= \sin\theta \cos\phi + \cos\theta \sin\phi \\ &= \sin(\phi+\theta),\end{align*}$$ from which it follows that $\phi+\theta = \pi/2 + 2\pi k$ for some integer $k$, hence $$\cos\theta = \cos\left(\frac{\pi}{2} + 2\pi k - \phi\right) = \cos\left(\frac{\pi}{2} - \phi\right) = \sin\phi = \frac{40}{41},$$ and we are done.
</s>