Question

If x=sec theta - tan theta and y=cosec theta + cot theta then prove that xy+1=y-x

Answer 1

Answer:

Hope it is answer

Answer 2

Answer:

Given: $x=\sec \theta- \tan \theta$ and  $y=\cosec \theta+ \cot \theta$

We have to prove:- xy + 1 = y - x

proof:

If $x=\sec \theta- \tan \theta$

Since, $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$

$x = \frac{1}{\cos \theta}- \frac{\sin \theta}{\cos \theta}$

Take L.C.M in above expression,

$x = \frac{1-\sin \theta}{\cos \theta}$

Since, $\sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$ and $\sin^{2} \frac{\theta}{2} + \cos^{2} \frac{\theta}{2} =1$

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

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

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

Dividing numerator and denominator by $\sin \frac{\theta}{2}$

$x = \frac{\cot^{2} \frac{\theta}{2} - 1}{\cot^{2} \frac{\theta}{2} + 1}$

Now, simplify the y,

$y=\cosec \theta+ \cot \theta$

Since, $\cosec \theta=\frac{1}{\sin \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$

$y = \frac{1}{\sin \theta}- \frac{\cos \theta}{\sin \theta}$

Take L.C.M in above expression,

$y = \frac{1-\cos \theta}{\sin \theta}$

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

$y=\cot \frac{\theta}{2}$

Now, replace $\cot \frac{\theta}{2}$ in $x = \frac{\cot^{2} \frac{\theta}{2} - 1}{\cot^{2} \frac{\theta}{2} + 1}$ by y

we get $x=\frac{y-1}{y+1}$

we need to prove xy+1=y-x

L.H.S

xy + 1

$y\frac{y-1}{y+1}+1$

$\frac{y^{2}-y+y+1}{y+1}$

$\frac{y^{2}+1}{y+1}$

R.H.S

y - x

$y -\frac{y-1}{y+1}$

$\frac{y^{2}+y-y+1}{y+1}$

$\frac{y^{2}+1}{y+1}$

L.H.S = R.H.S

Hence proved