<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] This has already been mentioned multiple times, but for completeness:
Let $\sin \theta = x, \cos \theta = y$. Then we have to solve:
$$9x+40y=41 \quad \quad x^2 + y^2 = 1$$
Now from the second equation, $(9x)^2 + 81y^2 = 81$, or $(41 - 40y)^2 + 81y^2 = 81$. This gives:
$$(40^2+9^2)y^2 - 2 \cdot 41 \cdot 40y + (41^2-9^2)=0$$
$$41^2 y^2 - 2 \cdot 41 \cdot 40 + 40^2 = 0$$
using the Pythagorean triple $9^2 + 40^2 = 41^2$, which factors as a perfect square:
$$(41y - 40)(41y - 40) = 0$$
$$\implies \cos \theta = \frac{40}{41}$$
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