Question

Prove that cosA/secA-tanA =1+sinA​

Answer 1

Answer:

$\frac{ \cos( \alpha ) }{ \sec( \alpha ) - \tan( \alpha ) } \\ \frac{ { \cos( \alpha ) }^{2} }{1 - \sin( \alpha ) } \\ \frac{1 - { \sin( \alpha ) }^{2} }{1 - \sin( \alpha ) } \\ 1 + \sin( \alpha )$

Hence proved...

Step-by-step explanation:

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Answer 2

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

$\boxed { \pink { sec^{2} A - tan^{2} A = 1 }}$

$= cosA( secA + tanA) \\= cosA secA + cosA tanA \\= cosA \times \frac{1}{cosA} + cosA \times \frac{sinA}{cosA} \\= 1 + sinA \\= RHS$

$Hence\: proved .$

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