Question
Prove that: $\sec A (1 - \sin A) (\sec A + \tan A) = 1$
✓
Answer
L.H.S.$= \sec A(1 - \sin A)(\sec A + \tan A)$
=$\frac{1}{\cos A} (1 - \sin A)( \frac{1}{\cos A}$ + $\frac{\sin A}{\cos A}$)
=$\frac{(1 - \sin A)}{\cos A}$($\frac{1 + \sin A}{\cos A}$)
= $\frac{(1-\sin A)(1+\sin A)}{\cos A \times \cos A}$
= $\frac{\left(1^{2}-\sin ^{2} A\right)}{\cos ^{2} A} .[ $Since$, (a - b ) (a + b ) = a^2 - b^2 ]$
= $\frac{\left(1-\sin ^{2} A\right)}{\cos ^{2} A}$
= $\frac{\cos^2 A}{\cos^2A}$
$= 1$
=RHS
Hence, proved.
=$\frac{1}{\cos A} (1 - \sin A)( \frac{1}{\cos A}$ + $\frac{\sin A}{\cos A}$)
=$\frac{(1 - \sin A)}{\cos A}$($\frac{1 + \sin A}{\cos A}$)
= $\frac{(1-\sin A)(1+\sin A)}{\cos A \times \cos A}$
= $\frac{\left(1^{2}-\sin ^{2} A\right)}{\cos ^{2} A} .[ $Since$, (a - b ) (a + b ) = a^2 - b^2 ]$
= $\frac{\left(1-\sin ^{2} A\right)}{\cos ^{2} A}$
= $\frac{\cos^2 A}{\cos^2A}$
$= 1$
=RHS
Hence, proved.
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