Question

cosA/1+sinA+1+sinA/cosA=2secA​

Answer 1

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$\tt\:LHS=\dfrac{cosA}{1+sinA}+\dfrac{1+sinA}{cosA}$

$\tt\longrightarrow\dfrac{{cos}^{2}A+(1+sinA)^2}{(1+sinA)\times\:cosA}$

$\tt\longrightarrow\dfrac{{cos}^{2}A+{(1)}^{2}+2\times\:1\times\:sinA+{(sinA)}^{2}}{(1+sinA)\times\:cosA}$

$\tt\longrightarrow\dfrac{{cos}^{2}A+1+2sinA+{sin}^{2}A}{(1+sinA)\times\:cosA}$

$\tt\longrightarrow\dfrac{{cos}^{2}A+{sin}^{2}A+1+2sinA}{(1+sinA)\times\:cosA}$

$\tt\longrightarrow\dfrac{1+1+2sinA}{(1+sinA)\times\:cosA}$

$\tt\longrightarrow\dfrac{2+2sinA}{(1+sinA)\times\:cosA}$

$\tt\longrightarrow\dfrac{2\cancel{(1+sinA)}}{\cancel{(1+sinA)}\times\:cosA}$

$\tt\longrightarrow\:2\times\:\dfrac{1}{cosA}$

$\tt\longrightarrow\:2\:secA$

$\tt\:LHS=RHS$

$\tt\:hence,proved$

Answer 2

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