Prove that tanA +sinA/tanA-sinA = secA …
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Posted by Aditya Narayan Panda 6 years, 9 months ago
- 1 answers
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Rashmi Bajpayee 6 years, 9 months ago
{tex}{{\tan {\rm{A}} + \sin {\rm{A}}} \over {\tan {\rm{A}} - \sin {\rm{A}}}}{/tex}
= {tex}{{{{\sin {\rm{A}}} \over {\cos {\rm{A}}}} + \sin {\rm{A}}} \over {{{\sin {\rm{A}}} \over {\cos {\rm{A}}}} - \sin {\rm{A}}}}{/tex}
= {tex}{{\sin {\rm{A}} + \cos {\rm{A}}.\sin {\rm{A}}} \over {\sin {\rm{A}} - \cos {\rm{A}}.\sin {\rm{A}}}}{/tex}
= {tex}{{\sin {\rm{A}}\left( {1 + \cos {\rm{A}}} \right)} \over {\sin {\rm{A}}\left( {1 - \cos {\rm{A}}} \right)}}{/tex}
= {tex}{{1 + \cos {\rm{A}}} \over {1 - \cos {\rm{A}}}}{/tex}
Proved.
Also, {tex}{{1 + \cos {\rm{A}}} \over {1 - \cos {\rm{A}}}}{/tex}
= {tex}{{1 + {1 \over {\sec {\rm{A}}}}} \over {1 - {1 \over {\sec {\rm{A}}}}}}{/tex}
= {tex}{{\sec {\rm{A}} + 1} \over {\sec {\rm{A}} - 1}}{/tex}
Proved.
2Thank You