Question

1+tanA/secA - SecA/1+tanA = 2sinAcosA/sinA+cosA



Please solve this guys

Answer 1

Answer:

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Answer 2

$\frac{1 + tana}{seca} - \frac{seca}{1 + tana} \\ \\ = \frac{1 + {tan}^{2}a + 2tana - {sec}^{2} a }{seca(1 + tana)} \\ \\ = \frac{2tana}{seca(1 + tana)} \\ \\ = \frac{2 \frac{sina}{cosa} }{ \frac{1}{cosa}( \frac{cosa + sina}{cosa} ) } \\ \\ = \frac{2sina}{cosa} \times \frac{ {cos}^{2} a}{sina + cosa} \\ \\ = \frac{2sinacosa}{sina + cosa}$