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প্রমাণ কর: $\frac{tanA}{secA+1}-\frac{secA-1}{tanA}=0$

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প্রমাণ কর: $\frac{tanA}{secA+1}-\frac{secA-1}{tanA}=0$

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Left Hand Side
$=\frac{tanA}{secA+1}-\frac{secA-1}{tanA}$

$=\frac{tan^2A-(secA+1)(secA-1)}{(secA+1)tanA}$

$=\frac{tan^2A-(sec^2A-1^2)}{(secA+1)tanA}$

$=\frac{tan^2A-(sec^2A-1)}{(secA+1)tanA}$

$=\frac{tan^2A-tan^2A}{(secA+1)tanA}$

$=\frac{0}{(secA+1)tanA}$

$=0$

$=$ Right Hand Side

[Proved]
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Rules Applied :

  • $a^2-b^2=(a+b)(a-b)$
  • $tan^2\theta=sec^2\theta-1$

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