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.
 A: Squaring should generally be avoided as it immediately introduces extraneous root
Write $9=r\sin y,40=r\cos y$ where $r>0$
Squaring & adding we get,  $r^2=41^2\implies r=41$ 
So, we have $r\cos(\theta-y)=41\iff\cos(\theta-y)=1=\cos0\implies\theta-y=2m\pi$ where $m$ is any integer
So, $\cos\theta=\cos(y+2m\pi)=\cos y=\dfrac{40}r $
A: I noted that $9,40$ and $41$ are pythagorean triple, i.e $9^2+40^2=41^2$. This fact can be used to solve the question easily. I'll show the solution for the general case:
$a\cos\theta+b\sin\theta=c,$ given that $a^2+b^2=c^2$     
$$a\cos\theta+b\sin\theta=c$$
$$b\sin\theta=c-a\cos\theta$$
Squaring both the sides:
$$b^2\sin^2\theta=c^2+a^2\cos^2\theta-2ca\cos\theta$$
Substituting $c^2$ with $a^2+b^2$ gives:
$$b^2\sin^2\theta=a^2+b^2+a^2\cos^2\theta-2ca\cos\theta$$
$$a^2+b^2\cos^2\theta+a^2\cos^2\theta-2ca\cos\theta=0$$
$$a^2+(a^2+b^2)\cos^2\theta-2ca\cos\theta=0$$
$$a^2+(c\cos\theta)^2-2a(c\cos\theta)=0$$
$$(a-c\cos\theta)^2=0$$
$$\cos\theta=\dfrac{a}{c}$$   
Substituting $c=41$ and $a=40$ will give the solution of the particular equation discussed in the question.
An interesting thing is that $\sin\theta=\sqrt{1-\dfrac{a^2}{c^2}}=\dfrac{b}{c}, $ , i.e the equation actually belongs to a right angled triangle having $a$ as its base, $b$ as its prependicular and $c$ as its hypotenuse. 
There is another method in which we divide both sides of the equation with $\sqrt{a^2+b^2}$ and then use the identity: 
$\sin (A+B)=\sin A \cos B+\cos A \cos B$   
This method will give the same results.
Converse
Consider a right angled triangle having base $a$, perpendicular $b$ and hypotenuse $c$. We have:
$$\cos\theta=\dfrac{a}{c}\ \  \text{and}\ \ \  \sin\theta=\dfrac{b}{c}$$
$$\implies a=c\cos\theta \ \ \ \text{and}\ \ \  b=c\sin\theta$$
By the pythagoras theorem:
$$a^2+b^2=c^2 $$
$$\implies\ c^2\cos^2\theta+c^2\sin^2\theta=c^2$$
$$\implies c\cos\theta \cdot c\cos\theta + c\sin\theta \cdot c\sin\theta=c^2$$
$$\implies a\cos\theta+b\sin\theta=c$$
A: Using Brahmagupta-Fibonacci Identity,
$$(9\sin\theta+40\cos\theta)^2+(40\sin\theta-9\cos\theta)^2=(40^2+9^2)(\sin^2\theta+\cos^2\theta)=41^2$$ as $41^2=40^2+9^2$
So, $40\sin\theta-9\cos\theta=0$ and $9\sin\theta+40\cos\theta=41$
Can you solve for $\cos\theta$
A: Look at the work that you have already done. You know that $1681=41^2$, that $3280=2\cdot 40\cdot 41$, and that $1600=40^2$. You found this out when you squared $41-40\cos\theta$. It should now be quite easy to solve the quadratic
$$1681\cos^2\theta−3280\cos\theta+1600=0$$
$$41^2\cos^2\theta−2(41)(40)\cos\theta−40^2=0$$
$$(41\cos\theta-40)^2=0$$
$$\dots$$
If you prefer to use the quadratic formula, remember to factor before you evaluate the square root.
$$\begin{array}{lll}
\cos\theta &=& \frac{3280\pm\sqrt{3280^2-4(1681)(1600)}}{2\cdot 1681}\\
&=&\frac{2\cdot40\cdot41\pm\sqrt{(2\cdot40\cdot41)^2-4(41^2)(40^2)}}{2\cdot 41\cdot41}\\
&=&\frac{2\cdot40\cdot41}{2\cdot 41\cdot41}\\
&=&\frac{40}{41}\\
\end{array}$$
A: 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.
A: 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:
$$(41y - 40)(41y - 40) = 0$$
$$\implies \cos \theta = \frac{40}{41}$$
