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(cosec theta -sin theta)(sec theta-cos theta)=1/(tan …

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(cosec theta -sin theta)(sec theta-cos theta)=1/(tan theta+cot theta)
  • 1 answers

Sia ? 4 years, 2 months ago

We have,
LHS = (cosec{tex}\theta{/tex} - sin{tex}\theta{/tex}) (sec{tex}\theta{/tex} - cos{tex}\theta{/tex})
{tex}\Rightarrow \quad \mathrm { LHS } = \left( \frac { 1 } { \sin \theta } - \sin \theta \right) \left( \frac { 1 } { \cos \theta } - \cos \theta \right){/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { 1 - \sin ^ { 2 } \theta } { \sin \theta } \times \frac { 1 - \cos ^ { 2 } \theta } { \cos \theta }{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { \cos ^ { 2 } \theta } { \sin \theta } \times \frac { \sin ^ { 2 } \theta } { \cos \theta }{/tex} (since, sin2A+cos2A =1)
{tex}\Rightarrow \quad \text { LHS } = \frac { \sin \theta \cos \theta } { 1 } = \frac { \sin \theta \cos \theta } { \sin ^ { 2 } \theta + \cos ^ { 2 } \theta } = \frac { 1 } { \frac { \sin ^ { 2 } \theta + \cos ^ { 2 } \theta } { \sin \theta \cos \theta } }{/tex}
{tex}\Rightarrow \quad L H S = \frac { 1 } { \frac { \sin ^ { 2 } \theta } { \sin \theta \cos \theta } + \frac { \cos ^ { 2 } \theta } { \sin \theta \cos \theta } } = \frac { 1 } { \tan \theta + \cot \theta } = R H S{/tex}

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