Inverse trigonometric functions without calculator, arcsin, arctan How can I evaluate this expression without the use of a calculator and only assuming i know the standard angles ($\pi/3, \pi/4, \pi/6$)?
\begin{align*}
\arcsin\left(\frac{5\sqrt{3}}{14}\right)-\arctan\left(\frac{1}{4\sqrt{3}}\right)
\end{align*}
 A: Let $\arcsin\dfrac{5\sqrt3}{14}=y,0<y<\dfrac\pi2$
$\sin y=?$
$\cos y=+\sqrt{1-\sin^2y}=?$
$\tan y=?$
$y=\arctan?$
Now $-\arctan(a)=\arctan(-a)$
Finally use Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
A: Hint: Draw a right triangle $\triangle ABC$ and label the lengths of two sides appropriately so that the base angle $\alpha=\arcsin\left(\frac{5\sqrt{3}}{14}\right)$ as in the diagram below. Then, by the Pythagorean Theorem, the remaining side must have length $11$.

Follow the example of the first triangle to appropriately label two sides of the second triangle $\triangle DEF$ so that the base angle at vertex $D$ is $\delta=\arctan\left(\frac{1}{4\sqrt{3}}\right)$. Use the Pythagorean Theorem to find the length of the third side.
Now the task is reduced to finding the angle $\alpha-\delta$. We know that both angles are positive acute angles and that $\alpha$ is the larger of the two, so we know that $\alpha-\delta$ is a positive acute angle and is one of the special angles which were mentioned in the problem.
Suggestion: Find $\sin(\alpha-\delta)$ using an identity which you studied and use the two triangles to find the values needed in the formula. You should find that $\sin(\alpha-\delta)$ is the sine of one of the special angles. Then you will have your answer.
