Sin(A-B)/cosA.cosB + sin (B-C)/cosB.cosC +sin (C-A)/cosC.cosA …

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Sin(A-B)/cosA.cosB + sin (B-C)/cosB.cosC +sin (C-A)/cosC.cosA =0

Posted by Jishan Khan 8 years, 4 months ago

CBSE > Class 11 > Mathematics

2 answers

Rashmi Bajpayee 8 years, 4 months ago

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

= {tex}{{\sin {\rm{A}}\cos {\rm{B}} - \cos {\rm{A}}\sin {\rm{B}}} \over {\cos {\rm{A}}\cos {\rm{B}}}} + {{\sin {\rm{B}}\cos {\rm{C}} - \cos {\rm{B}}\sin {\rm{C}}} \over {\cos {\rm{B}}\cos {\rm{C}}}} + {{\sin {\rm{C}}\cos {\rm{A}} - \cos {\rm{C}}\sin {\rm{A}}} \over {\cos {\rm{C}}\cos {\rm{A}}}}{/tex}

= {tex}\eqalign{ & {{\sin {\rm{A}}\cos {\rm{B}}\cos {\rm{C}} - \cos {\rm{A}}\sin {\rm{B}}\cos {\rm{C}} + \cos {\rm{A}}\sin {\rm{B}}\cos {\rm{C}} - \cos {\rm{A}}\cos {\rm{B}}\sin {\rm{C}} + \sin {\rm{C}}\cos {\rm{A}}\cos {\rm{B}} - \cos {\rm{B}}\cos {\rm{C}}\sin {\rm{A}}} \over {\cos {\rm{A}}\cos {\rm{B}}\cos {\rm{C}}}} \cr & \cr} {/tex}

= {tex}{0 \over {\cos {\rm{A}}\cos {\rm{B}}\cos {\rm{C}}}}{/tex}

= 0

Hence proved

2Thank You

Jishan Khan 8 years, 4 months ago

Prove 0

3Thank You

ANSWER