Question

1÷1+sinA + 1÷1-sinA​

Answer 1

Step-by-step explanation:

$\frac{1}{1+sinA} + \frac{1}{1-sinA} = \frac{1-sinA+1+sinA}{(1+sinA)(1-sinA)} = \frac{2}{1-sin^2A} = \frac{2}{cos^2A} = 2sec^2A$

Answer 2

Answer:

$\large 2\sec^2A$

Step-by-step explanation:

We have to simplify

$\large \dfrac{1}{1+\sin A}+\dfrac{1}{1-\sin A} \\\\\\\large \implies\dfrac{1}{1+\sin A}+\dfrac{1}{1-\sin A}\\\\\\\large \implies\dfrac{1-\sin A+1\sin A}{(1+\sin A)(1-\sin A)}\\\\\\\implies\dfrac{1-\sin A+1\sin A}{1-\sin^2A}\\\\\\\implies\dfrac{2}{1-\sin^2A}$

We know that

$\large 1-\sin^2A=\cos^2A$

Put value here

$\large \implies\dfrac{2}{\cos^2A}\\\\\\\large \implies2\times\dfrac{1}{\cos^2A}$

$\large \text{We know that \dfrac{1}{\cos^2A} can be written as \sec^2A }$

So write it here we get

$\large \implies2\times\dfrac{1}{\cos^2A}\\\\\\\large \implies2\times\sec^2A\\\\\\\large \implies2\sec^2A$

Thus we get answer.