Verify Trigonometric Identities $1/\sin50^\circ + √3/\cos50^\circ=4$
I have tried it as:
LHS
$(\cos50+√3 \sin50)/\sin50\cos50$
$(2\cos50+2√3 \sin50)/2\sin50\cos50$
$(2\cos50+2√3  \sin50)/(\sin100)$
Now, whats next??
 A: Hints:
$a\sin \theta + b \cos \theta = \sqrt {a^2 + b^2}\sin(\theta + \arctan {\frac ba})$
$\sin \theta = \sin (180^{\circ} - \theta)$
A: Continuing...
$$\frac{2\cos50+2\sqrt{2}\sin50}{\sin100}=4\times\frac{\frac12\cos50+\frac{\sqrt{3}}{2}\sin50}{\sin80}$$
$$=4\times\frac{\sin30\cos50+\cos30\sin50}{\sin80}$$
$$=4\times\frac{\sin80}{\sin80}$$
$$=4$$
A: \begin{align}
& \frac 1 {\sin50^\circ} + \frac{\sqrt 3}{\cos50^\circ} = 2 \left( \frac {1/2} {\sin50^\circ} + \frac{\sqrt 3/2}{\cos50^\circ} \right) = 2\left( \frac{\sin30^\circ}{\sin50^\circ} + \frac{\cos 30^\circ}{\cos50^\circ} \right) \\[10pt]
= {} &  2\cdot \frac{\sin30^\circ\cos50^\circ + \cos30^\circ\sin30^\circ}{\sin50^\circ\cos50^\circ} = 4\cdot \frac{\sin(30^\circ+50^\circ)}{2\sin50^\circ\cos50^\circ} \\[10pt]
= {} & 4\cdot \frac{\sin(30^\circ+50^\circ)}{\sin(2\cdot50^\circ)} = 4\cdot\frac{\sin80^\circ}{\sin100^\circ}
\end{align}
Now recall why $\sin80^\circ = \sin100^\circ$.
A: $$\dfrac{\sin3A}{\sin x}+\dfrac{\cos3A}{\cos x}=2\cdot\dfrac{\sin(3A+x)}{\sin2x}$$
If $\sin2x=\sin(3A+x),2x=n180^\circ+(-1)^n3(A+x)$ where $n$ is any integer
If $n=2m$(even), $x=m360^\circ+3A$
If $n=2m+1$(odd)  $x=\dfrac{(2m+1)180^\circ-3A}3=(2m+1)60^\circ-A$
Here $A=10^\circ\implies x=120^\circ m+50^\circ$
