Question

Under root 1+cos theta/1-cos theta=cosec theta+cot theta then the quadrant in which theta lies is

Answer 1

GIVEN :–

$\\ \: \blacktriangleright { \bold{ \sqrt{ \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) } } = cosec( \theta) + \cot( \theta) }}$

TO FIND :–

• The quadrant in which $\:\: \theta \:\:$ lies is = ?

SOLUTION :–

$\\ \: \implies { \bold{ \sqrt{ \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) } } = cosec( \theta) + \cot( \theta) }}$

• We should write this as –

$\\ \: \implies { \bold{ \sqrt{ \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) } \times \frac{1 + \cos( \theta) }{ 1 + \cos( \theta) } } = cosec( \theta) + \cot( \theta) }}$

$\\ \: \implies { \bold{ \sqrt{ \dfrac{ \{1 + \cos( \theta) \}^{2} }{ {(1)}^{2} - \cos ^{2} ( \theta) }} = cosec( \theta) + \cot( \theta) }}$

$\\ \: \implies { \bold{ \sqrt{ \dfrac{ \{1 + \cos( \theta) \}^{2} }{1 - \cos ^{2} ( \theta) }} = cosec( \theta) + \cot( \theta) }}$

$\\ \: \implies { \bold{ \sqrt{ \dfrac{ \{1 + \cos( \theta) \}^{2} }{\sin^{2} ( \theta) }} = cosec( \theta) + \cot( \theta) }}$

$\\ \: \implies { \bold { \sqrt{ \left \{ \dfrac{1 + \cos( \theta)}{\sin ( \theta) } \right \}^{2} } = cosec( \theta) + \cot( \theta) }}$

$\\ \: \implies { \bold { \sqrt{ \{ cosec( \theta) + \cot( \theta) \}^{2} } = cosec( \theta) + \cot( \theta) }}$

$\\ \: \implies { \bold { \pm \: { \{ cosec( \theta) + \cot( \theta) \} } = cosec( \theta) + \cot( \theta) }}$

• To satisfy the equation —

$\\ \: \blacktriangleright \: { \bold {cosec( \theta) \: \: and \: \: \cot( \theta) \: \:must \: \: be \: \: positive.}}$

• We also know that –

$\\ \: \blacktriangleright \: { \bold {cosec( \theta) \: \: positive \: \: when \: \: \theta \: \: in \: \: first \:\: and \: \: second \: \: quadrant .}}$

• And –

$\\ \: \blacktriangleright \: { \bold { \cot( \theta) \: \: positive \: \: when \: \: \theta \: \: in \: \: first \:\: and \: \: third \: \: quadrant .}}$

• Hence , $\:\: \theta \:\:$ in first quadrant.