Question

If cosθ+sinθ=1,then prove that cosθ−sinθ=±1

Answer 1

GIVEN :–

$\\ { \bold{ \cos( \theta) + \sin( \theta) = 1 }}$

TO PROVE :–

$\\ { \bold{ \cos( \theta) - \sin( \theta) = \pm \: 1 }}$

SOLUTION :–

$\\ \implies { \bold{ \cos( \theta) + \sin( \theta) = 1 }}$

• Square on both sides –

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

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

$\\ \implies { \bold{ 1 + 2 \cos( \theta) \sin( \theta) = 1 \: \: \: \: \: [ \: \because \: \cos^{2} ( \theta) + \sin^{2} ( \theta) = 1]}}$

$\\ \implies { \bold{ 2 \cos( \theta) \sin( \theta) = 0}}$

$\\ \implies { \bold{ \cos( \theta) \sin( \theta) = 0 \: \: \: \: - - - eq.(1)}}$

• Now Let's take L.H.S. –

$\\ \: \: = \: \: { \bold{ \cos( \theta) - \sin( \theta)}}$

• We should write this as –

$\\ \: \: = \: \: { \bold{ \pm \sqrt{[ \cos( \theta) - \sin( \theta) ] ^{2} }}}$

$\\ \: \: = \: \: { \bold{ \pm \sqrt{\cos^{2} ( \theta) + \sin^{2} ( \theta) - 2 \cos( \theta) \sin( \theta) }}}$

• Using eq.(1) –

$\\ \: \: = \: \: { \bold{ \pm \sqrt{\cos^{2} ( \theta) + \sin^{2} ( \theta) - 2 (0)}}}$

$\\ \: \: = \: \: { \bold{ \pm \sqrt{\cos^{2} ( \theta) + \sin^{2} ( \theta) }}}$

$\\ \: \: = \: \: { \bold{ \pm \sqrt{1}}}$

$\\ \: \: = \: \: { \bold{ \pm \: 1}}$

$\\ \: \: = \: \: { \bold{ \: R.H.S.}}$

$\\ \\ \: \: \: \: \: \: \: \: \: { \bold{ \underbrace {Hence \: \: \: proved}}}$