1 টি উত্তর পাওয়া গেছে

Left Hand Side

$=\frac{tanA}{secA+1}-\frac{secA-1}{tanA}$

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

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

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

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

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

$=0$

$=$ Right Hand Side

[ Proved ]

কমেন্টস

Rules Applied :

  • $tan^2 \theta=sec^2 \theta - 1$