Question

cosA÷1+sinA+1+sinA÷cosA=2secA​

Answer 1

LHS =cosA/(1+sinA) +(1+sinA)/cosA

={cos²A +(1+sinA)²}/cosA.(1+sinA)

={cos²A+1+sin²A+2sinA}/cosA.(1+sinA)

use sin²∅ +cos²∅ =1

=(1+1+2sinA)/cosA(1+sinA)

=2(1+sinA)/cosA(1+sinA)

=2/cosA

=2secA = RHS

according to this u can get ur answer

mark as brainliest if ur indian

Answer 2

Answer:

$\frac{cosa}{1 + sina} + \frac{1 + sina}{cosa} \\ \\ = \frac{ {cos}^{2}a + ( {1 + sina})^{2} }{(1 + sina)cosa} \\ \\ = \frac{ {cos}^{2}a + 1 + {sin}^{2} a + 2sina }{(1 + sina)cosa} \\ \\ = \frac{2 + 2sina}{(1 + sina)cosa} \\ \\ = \frac{2}{cosa} \\ \\ = 2seca$