If $\frac{a}{\sin{A}}=\frac{b}{\cos{A}}$, show that $\sin{A}\cos{A}=\frac{ab}{a^2+b^2}$

Question

I don't know how to go about solving this, I think I need to use $\sin^2\theta+\cos^2\theta=1$, but I'm not sure how to go about this.

The closest I managed to get was:

$$\frac{a}{\sin{A}}=\frac{b}{\cos{A}}$$ $$a\cos{A}-b\sin{A}=0$$ $$a^2\cos^2{A}+b^2\sin^2{A}-2ab\cos{A}\sin{A}=0$$ $$\cos{A}\sin{A}=\frac{a^2\cos^2{A}+b^2\sin^2{A}}{2ab}$$

Which seems to vaguely resemble what I need, but I'm not sure of the final steps.

Answer 1

Notice that $$ \sin A \cos A = \frac{\sin A \cos A}{\sin^2 A + \cos^2 A} = \frac{\tan A}{\tan^2 A + 1} $$ It should be ease for you to continue.

Answer 2

Hint: just substitute $a=b\frac {\sin \, A}{\cos \, A}$ in $\frac {ab} {a^{2}+b^{2}}$ and simplify.

Answer 3

If $\frac{a}{\sin A} = \frac b{\cos A}$, that is a right angled triangle of one angle $A$, opposite side $a$, adjacent side $b$ and hypotenuse $\sqrt {a^2 + b^2}$.

Then $\sin A$ is really a scaled version of $a$ by $\dfrac 1{\sqrt {a^2 + b^2}}$, and $\cos A$ is a scaled version of $b$ by the same factor.

Answer 4

Note that you have $$\frac{a}{\sin{A}}=\frac{b}{\cos{A}}$$ which is equivalent to $$ \tan A = \frac {a}{b}$$

Now $$\sin A \cos A = (1/2) \sin (2A) = \frac {\tan A}{1+ \tan^2 A} = \frac {ab}{a^2 + b^2}$$

Answer 5

Hint

$$\dfrac a{\sin A}=\dfrac b{\cos A}=\pm\sqrt{\dfrac{a^2+b^2}{\sin^2A+\cos^2A}}$$

Observe that the first two ratios have the same sign