Question

Prove that :
(sin A+ sec A)²+(cos A+codec A)²
=(1+ sec A codec A)²

Answer 1

$\huge{\boxed{\mathtt{\red{Solution:-}}}}$

$\bf \: L.H.S.$

$\sf = \: (sin \: A \: + \: sec \: A) {}^{2} + (cos \: A \: + \: cosec \: A) {}^{2}$

$\sf = \bigg(sin \: A \: + \frac{1}{cos \: A} \bigg) {}^{2} + \bigg( cos \: A + \frac{1}{sin \: A} \bigg) {}^{2}$

$\sf = \: \bigg( \frac{sin \: A \: cos \: A + }{cos \: A} \bigg) {}^{2} + \bigg( \frac{sin \: A \: cos \: A \:+1}{sin \: A} \bigg) {}^{2}$

$\sf = \: \frac{(sin \: A \: cos \: A \: + 1) {}^{2} }{cos {}^{2} \: A} + \frac{(sin \: A \: cos \: A + 1) {}^{2} }{sin {}^{2}A}$

$\sf \: Taking \: \: (sin \: A \: cos \: A \: + 1) {}^{2} \: common,$

$= \sf \: (sin \: A\: cos \: A \: + 1) {}^{2} \bigg( \frac{1}{cos {}^{2}A} + \frac{1}{sin {}^{2} A} \bigg)$

$= \sf \: (sin \: A \: cos \: A + 1) {}^{2} \bigg( \frac{sin {}^{2} A + cos {}^{2} A}{cos {}^{2} A \: sin {}^{2} A} \bigg)$

$= \: \sf \: (sin \: A \: cos \: A + 1) {}^{2} \: \frac{1}{(cos \: A \: sin \: A)}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \: \: ∵ \: sin {}^{2} \: A + cos {}^{2} A = 1$

$= \sf \bigg( \frac{sin \: A \: cos \: A + 1}{cos \: A \: sin \: A} \bigg) {}^{2} \: \: \: \: \: \: \: \bigg | \: ∵\frac{ a {}^{2} }{b {}^{2} } = \bigg( \frac{a}{b} \bigg) {}^{2}$

$= \: \sf \bigg(\frac{sin \: A \: cos \: A }{cos \: A \: cos \: A} + \frac{1}{cos \: A \: sin \: A} \bigg) {}^{2}$

$\sf = \: \bigg(1 + \frac{1}{cos \: A} \frac{1}{sin \: A} \bigg) {}^{2}$

$\sf = \: (1 + sec \: A \: cosec \: A) {}^{2} = \: R.H.S$