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(1+cotA+tanA)(sinA-cosA)=secA/cosec2A-cosecA/sec2A  

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(1+cotA+tanA)(sinA-cosA)=secA/cosec2A-cosecA/sec2A

 

  • 2 answers

Nidhi Pal 4 years, 1 month ago

I can't understand

Rashmi Bajpayee 6 years, 9 months ago

L.H.S.

{tex}\left( {1 + \cot {\rm{A}} + \tan {\rm{A}}} \right)\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right){/tex}

= {tex}\left( {1 + {{\cos {\rm{A}}} \over {\sin {\rm{A}}}} + {{\sin {\rm{A}}} \over {\cos {\rm{A}}}}} \right)\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right){/tex}

= {tex}{{\left( {\sin {\rm{A}} - \cos {\rm{A}}} \right)\left( {\sin {\rm{AcosA}} + {{\sin }^2}{\rm{A}} + {{\cos }^2}{\rm{A}}} \right)} \over {\sin {\rm{AcosA}}}}{/tex}

= {tex}{{{{\sin }^3}{\rm{A}} - {{\cos }^3}{\rm{A}}} \over {\sin {\rm{AcosA}}}}{/tex}

= {tex}{{{{\sin }^2}{\rm{A}}} \over {{\rm{cosA}}}} - {{{{\cos }^2}{\rm{A}}} \over {\sin {\rm{A}}}}{/tex}

= {tex}{1 \over {{\rm{cosA}}}}.{{{{\sin }^2}{\rm{A}}} \over 1} - {1 \over {{\rm{sinA}}}}.{{{{\cos }^2}{\rm{A}}} \over 1}{/tex}

= {tex}{{\sec {\rm{A}}} \over {\cos e{c^2}{\rm{A}}}} - {{\cos ec{\rm{A}}} \over {{{\sec }^2}{\rm{A}}}}{/tex}

= R.H.S.

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