Question

If $\sin \theta +\cos \theta =m$ and $\sec \theta +\csc \theta =n$, then prove that $n(m^2-1)=2m$

Answer 1

Given:- $\sin \theta +\cos \theta =m\sec \theta +\csc \theta =n$

To prove:- $n(m^2-1)=2m$

Proof:-

$\sec \theta +\csc \theta =n$

$\Rightarrow \frac{1}{\cos \theta }+\frac{1}{\sin \theta }=n\Rightarrow \frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }=n\Rightarrow \frac{m}{n}=\sin \theta \cos \theta$

Now,

$n(m^2-1)=2m$

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

${(\sin \theta +\cos \theta )}^2-1=2\sin \theta \cos \theta$

${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta -1=2\sin \theta \cos \theta$

$1-1=2\sin \theta -\cos \theta -2\sin \theta \cos \theta$

$0=0$

L.H.S. $=$ R.H.S.

Hence proved.