The value of `[(secA+tanA)(1-sinA)]` is equal to `tan^2A(b)sin^2A` (c) `cosA` (d) `sinA`

Explanation

`(\sec {A}+\tan {A})(1-\sin {A}) `


`=\left(\frac{1}{\cos {A}}+\frac{\sin {A}}{\cos {A}}\right)(1-\sin {A}) `


`=\left(\frac{1+\sin {A}}{\cos {A}}\right)(1-\sin {A}) `


`=\frac{(1+\sin {A})(1-\sin {A})}{\cos {A}}= `


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