Question

Prove that
(tano/seco-1) - (sino/
1+coso)
= 2 coto

Answer 1

LHS:

$(\frac{tan \: a}{sec \: a - 1} ) - ( \frac{sin \: a}{1 + cos \: a} ) = 2 \: cot \: a \\ ( \frac{ \frac{sin \: a}{cos \: a} }{ \frac{1 - cos \: a}{cos \: a} } ) - ( \frac{sin \: a}{1 + cos \: a} ) \\ ( \frac{sin \: a}{1 - cos \: a} ) - ( \frac{sin \: a}{1 + cos \: a} ) \\ ( \frac{sin \: a(1 + cos \: a) - sin \: a(1 - cos \: a)}{(1 - cos \: a)(1 + cos \: a)} \\ \frac{sin \: a + sin \: a \: cos \: a - sin \: a + sin \: a \: cos \: a}{ {sin}^{2}a } \\ \frac{2sin \: a \: cos \: a}{ {sin}^{2} a} \\ \\ 2 \frac{cos \: a}{sin \: a} \\ \\ 2cot \: a$

LHS = RHS

Hence proved

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