Question

if cos73=m/n show that cosec73-sec17=m/n√n2-m2

Answer 1

your question is wrong.

I think your question may be

If cos73=m/n show that

$cosec\,73-sec\,73=\frac{mn-n\sqrt{n^2-m^2}}{m\sqrt{n^2-m^2}}$

Given:

$cos\,73=\frac{m}{n}$

We know that

$sin^2\,73=1-cos^2\,73$

$sin^2\,73=1-\frac{m^2}{n^2}$

$sin^2\,73=\frac{n^2-m^2}{n^2}$

$sin\,73=\frac{\sqrt{n^2-m^2}}{n}$

$\implies\bf\,cosec\,73=\frac{n}{\sqrt{n^2-m^2}}$

Also,

$sec\,73=\frac{n}{m}$

Now,

$cosec\,73-sec\,73$

$=\frac{n}{\sqrt{n^2-m^2}}-\frac{n}{m}$

$=\frac{mn-n\sqrt{n^2-m^2}}{m\sqrt{n^2-m^2}}$

Answer 2

$\csc73-\cos17=\dfrac{m}{n\sqrt{n^2-m^2}}$, proved.

Step-by-step explanation:

We have,

$\cos73=\dfrac{m}{n}$

To prove that, $\csc73-\cos17=\dfrac{m}{n\sqrt{n^2-m^2}}$.

∴ $\cos73=\dfrac{m}{n}$

⇒ $\sin73=\dfrac{\sqrt{n^2-m^2}}{n}=\dfrac{p}{h}$ [ ∵ $\cos(90-A)=\sin A$]

Where, b = base and h = hypotaneous

∴ Perpendicular, $p=\sqrt{h^{2}-b^{2}}$

$=\sqrt{n^{2}-m^{2}}$

∴ $\csc73=\dfrac{h}{p} =\dfrac{n}{\sqrt{n^2-m^2}}$

L.H.S. = $\csc73-\cos17$

= $\csc73-\sin73$

= $\dfrac{n}{\sqrt{n^2-m^2}}-\dfrac{\sqrt{n^2-m^2}}{n}$

= $\dfrac{m}{n\sqrt{n^2-m^2}}$

= R.H.S., proved.