Question

(secO + tanO)^2=
(cosecO + 1) (cosecO- 1)​

Answer 1

$LHS = (secO + tanO)^{2} \\= \Big(\frac{1}{cos O} + \frac{sin O}{cos O} \Big)^{2} \\= \Big( \frac{(1+sin O)}{cos O}\Big)^{2} \\= \frac{(1+sin O)^{2}}{cos^{2} O } \\= \frac{(1+sin O)^{2}}{(1- sin^{2} O )} \\= \frac{(1+sin O)^{2}}{(1+sin O)( 1 - sin O)} \\=\frac{1+sin O}{1 - sin O }$

/* Dividing numerator and denominator by sin O , we get */

$= \frac{ \frac{1}{sin O} + 1 }{\frac{1}{sin O } - 1 } \\= \frac{Cosec O + 1}{Cosec O - 1 } \\= RHS$

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