Question

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Answer 1

To prove -

$\frac{\sin \theta }{[ \sec \theta + tan \theta - 1 ]} + \frac{\cos \theta }{[ \cosec \theta + \cot \theta - 1 ] } = 1$

Taking LHS

$\frac{\sin \theta }{[\sec \theta + \tan \theta - 1 ]} + \frac{\cos \theta }{[\cosec \theta + \cot \theta - 1 ]}$

Converting all values in term of sin and cos

$\frac{\sin \theta }{ \frac{1}{\cos \theta } + \frac{\sin \theta}{\cos \theta} -1} + \frac{\cos \theta}{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta } -1 }$

$\frac{\sin \theta . \cos \theta }{1 + \sin \theta - \cos \theta } + \frac{\sin \theta . \cos \theta }{1 - \sin \theta + \cos \theta }$

Taking sin theta . cos theta common -

$\sin \theta . \cos \theta [ \frac{1}{1 + \sin \theta - \cos \theta } + \frac{1 }{1 + \cos \theta - \sin \theta }]$

$\sin \theta . \cos \theta [ \frac{1 - \sin \theta + \cos \theta + 1+ \sin \theta - \cos \theta }{ [1 + \sin \theta - \cos \theta ] [ 1 + \cos \theta - \sin \theta ] } ]$

We know that [a+b][a-b] = $a^{2} - b^{2}$

$\frac{2 \sin \theta. \cos \theta }{{1}^{2} - {\sin \theta - \cos \theta }^{2} }$

$\frac{2 \sin \theta . \cos \theta }{1 - [ \sin^{2} \theta + \cos^{2} \theta - 2\cos \theta . \sin \theta ] }$

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

$\frac{2 \sin \theta \cos \theta }{ 2 \sin \theta \cos \theta } = 1$

Hence proved

Answer 2

Question :–

$\\ \: \: { \bold{prove \: \: that : \left[ \dfrac{ \sin( \theta) }{ \sec( \theta) + \tan( \theta) - 1 } \right] + \left[ \dfrac{ \cos( \theta) }{ cosec( \theta) + \cot( \theta) - 1 } \right] = 1}}$

ANSWER :–

• Let's take L.H.S. –

$\\ \: \: { \bold{ = \left[ \dfrac{ \sin( \theta) }{ \sec( \theta) + \tan( \theta) - 1 } \right] + \left[ \dfrac{ \cos( \theta) }{ cosec( \theta) + \cot( \theta) - 1 } \right] }}$

• We know that –

$\\ \longrightarrow{ \bold { \sec( \theta) = \dfrac{1}{ \cos( \theta) } \: \: and \: \: cosec( \theta) = \dfrac{1}{ \sin( \theta) }}}$

$\\ \longrightarrow{ \bold { \tan( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) } \: \: and \: \: \cot( \theta) = \dfrac{ \cos( \theta) }{ \sin( \theta) }}}$

• So that –

$\\ \: \: { \bold{ = \left[ \dfrac{ \sin( \theta) }{ \dfrac{1}{ \cos( \theta) } + \dfrac{ \sin( \theta) }{ \cos( \theta)} - 1 } \right] + \left[ \dfrac{ \cos( \theta) }{ \dfrac{1}{ \sin( \theta) } + \dfrac{ \cos( \theta) }{ \sin( \theta) } - 1 } \right] }}$

$\\ \: \: { \bold{ = \left[ \dfrac{ \sin( \theta) \times \cos( \theta) }{ 1 + \sin( \theta) - \cos( \theta) } \right] + \left[ \dfrac{ \cos( \theta) \times \sin( \theta) }{ 1 + \cos( \theta) - \sin( \theta) } \right] }}$

$\\ \: \: { \bold{ = (\sin( \theta) \times \cos( \theta))\left[ \dfrac{ 1 }{ 1 + \sin( \theta) - \cos( \theta) } + \dfrac{ 1 }{ 1 + \cos( \theta) - \sin( \theta) } \right] }}$

$\\ \: \: { \bold{ = (\sin( \theta) \times \cos( \theta))\left[ \dfrac{ 1 + \sin( \theta) - \cos( \theta) + 1 + \cos( \theta) - \sin( \theta) }{( 1 + \sin( \theta) - \cos( \theta) ).( 1 + \cos( \theta) - \sin( \theta) ) } \right] }}$

$\\ \: \: { \bold{ = (\sin( \theta) \times \cos( \theta))\left[ \dfrac{ 2 }{( 1 + \sin( \theta) - \cos( \theta) ).[ 1 - ( \sin( \theta ) - \cos( \theta) )] } \right] }}$

$\\ \: \: { \bold{ = (\sin( \theta) \times \cos( \theta))\left[ \dfrac{ 2 }{[ (1) ^{2} - ( \sin( \theta ) - \cos( \theta) )^{2} ] } \right] \: \: \: \: \:[ \: \because \: \:(a - b)(a + b) = {a}^{2} - {b}^{2} ] }}$

$\\ \: \: { \bold{ = ( \sin( \theta) \times \cos( \theta)) \left[ \dfrac{ 2 }{ 1 - { \sin}^{2}( \theta) - { \cos }^{2} ( \theta) + 2 \sin( \theta) \cos( \theta) } \right] }}$

$\\ \: \: { \bold{ = ( \sin( \theta) \times \cos( \theta)) \left[ \dfrac{ 2 }{ 1 - ( { \sin}^{2}( \theta) + { \cos }^{2} ( \theta) )+ 2 \sin( \theta) \cos( \theta) } \right] }}$

$\\ \: \: { \bold{ = ( \sin( \theta) \times \cos( \theta)) \left[ \dfrac{ 2 }{ 1 - 1+ 2 \sin( \theta) \cos( \theta) } \right] }}$

$\\ \: \: { \bold{ = ( \sin( \theta) \times \cos( \theta)) \left[ \dfrac{ \cancel2 }{ \cancel2 \sin( \theta) \cos( \theta) } \right] }}$

$\\ \: \: { \bold{ = \left[ \cancel\dfrac{ \sin( \theta) \cos( \theta) }{ \sin( \theta) \cos( \theta) } \right] }}$

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

$\\ \: \: { \bold{ = R.H.S. \: \: (Hence \: \: proved) }}$

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