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sinAsin2A+sin3Asin6A+sin4Asin13A/sinAcos2A+sin3Acos6A+sin4Acos13A = tan9A

  1. sinAsin2A+sin3Asin6A+sin4Asin13A/sinAcos2A+sin3Acos6A+sin4Acos13A = tan9A
    1. answers icon 2 answers
    2. Kewal asked by Kewal
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  2. SinAsin2A+sin3Asin6A/sinAcos2A+sin3Acos6A
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    2. HIRDESH asked by HIRDESH
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  3. PROVE THAT sin3Asin6A+sinAsin2A/sin3Acos6A+sinAcos2A
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    2. VALENTINE asked by VALENTINE
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