If cosecA+secA=cosecB+secB, then show that tanAtanB=cot(A+B)/2 …
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Posted by Himanshu Goyal 6 years, 9 months ago
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Rashmi Bajpayee 6 years, 9 months ago
{tex}\cos ec{\rm{A}} + \sec {\rm{A}} = \cos ec{\rm{B}} + \sec {\rm{B}}{/tex}
=> {tex}\cos ec{\rm{A}} - \cos ec{\rm{B}} = \sec {\rm{A}} - \sec {\rm{B}}{/tex}
=> {tex}{1 \over {\sin {\rm{A}}}} - {1 \over {\sin {\rm{B}}}} = {1 \over {\cos {\rm{A}}}} - {1 \over {\cos {\rm{B}}}}{/tex}
=> {tex}{{\sin {\rm{B}} - \sin {\rm{A}}} \over {\sin {\rm{A}}{\rm{.}}\sin {\rm{B}}}} = {{\cos {\rm{B}} - \cos {\rm{A}}} \over {\cos {\rm{A}}{\rm{.cos B}}}}{/tex}
=> {tex}{{2\sin {{{\rm{B}} - {\rm{A}}} \over 2}\cos {{{\rm{B}} + {\rm{A}}} \over 2}} \over {\sin {\rm{A}}{\rm{.}}\sin {\rm{B}}}} = {{2\sin {{{\rm{B}} - {\rm{A}}} \over 2}\sin {{{\rm{B}} + {\rm{A}}} \over 2}} \over {\cos {\rm{A}}{\rm{.cos B}}}}{/tex}
=> {tex}{{\cos {{{\rm{B}} + {\rm{A}}} \over 2}} \over {\sin {{{\rm{B}} + {\rm{A}}} \over 2}}} = {{\sin {\rm{A}}{\rm{.}}\sin {\rm{B}}} \over {\cos {\rm{A}}{\rm{.cos B}}}}{/tex}
=> {tex}\cot {{{\rm{B}} + {\rm{A}}} \over 2} = \tan {\rm{A}}{\rm{.}}\tan {\rm{B}}{/tex}
=> {tex}\tan {\rm{A}}{\rm{.}}\tan {\rm{B = }}\cot {{{\rm{B}} + {\rm{A}}} \over 2}{/tex}
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