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ashna

  • 2 years ago

If tan A - tan B = x and cot B - cot A =Y , pt COT (a-b) = 1/X + 1/Y

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  1. ashna
    • 2 years ago
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    @akash123

  2. akash123
    • 2 years ago
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    sin(A-B)/ cosA*cosB =1/x from Tan A - tan B=x

  3. ashna
    • 2 years ago
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    okay !

  4. ashna
    • 2 years ago
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    then ?

  5. akash123
    • 2 years ago
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    and sin(A-B)/ sinA* SinB=y from cotB- cotA =y sin(A-B)/ cosA*cosB =x from Tan A - tan B=x tanA*tanB= x/y

  6. sauravshakya
    • 2 years ago
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    @akash123 sin(A-B)/ cosA*cosB =x from Tan A - tan B=x

  7. akash123
    • 2 years ago
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    cot(A-B) = [ cot A* cot B+1] / ( cot B - cot A)

  8. akash123
    • 2 years ago
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    yes...thanks...I have corrected it later

  9. akash123
    • 2 years ago
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    sin(A-B)/ sinA* SinB=y from cotB- cotA =y sin(A-B)/ cosA*cosB =x from Tan A - tan B=x divide 2nd by 1st then u'll get tanA*tanB= x/y

  10. ashna
    • 2 years ago
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    yes i understood that !

  11. akash123
    • 2 years ago
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    now use the identity cot(A-B) = [ cot A* cot B+1] / ( cot B - cot A) and substitute the value of cot A* cot B and (cot B- cot A) then u'll get the answer

  12. akash123
    • 2 years ago
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    another way @sauravshakya write the reciprocal f 1st and 2nd equation and add both then simplify it.

  13. sauravshakya
    • 2 years ago
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    Yep... I would go with that....

  14. ashna
    • 2 years ago
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    okay got it :) thank you !

  15. akash123
    • 2 years ago
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    :)

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