Question

Prove it

$\frac{sin \: a + cos \: a}{sina \: - cos \: a} = \sqrt{ \frac{1 + sin2a}{1 - sin2a} }$

Answer 1

cosA +sinA= √2cosA

=> sinA= √2cosA- cosA

=> sinA = cosA(√2 -1)

Now multiplying both side by ( √2+1)

=>sinA(√2+1) = cosA(√2–1)(√2+1)

=>√2sinA + sinA=cosA(2–1)

=> cosA-sinA= √2sinA (Answer)

Answer 2

$\bold{\huge{Hay!!}}$


$\bold{Dear\:user!!}$



$\bold{\underline{Question-}}$


Prove it

$\frac{sin \: a + cos \: a}{sina \: - cos \: a} = \sqrt{ \frac{1 + sin2a}{1 - sin2a} }$



$\bold{\underline{Answer-}}$


LHS = Prove it

$\frac{sin \: a + cos \: a}{sina \: - cos \: a}$


$= > \sqrt{ \frac{(sin \: a \: + cos \: a) {}^{2} }{(sin \: a \: - cos \: a) {}^{2} } } \\ \\ \\ \\ = > \sqrt{ \frac{ {sin}^{2} a + {cos}^{2} a + 2sin \: a \: cosa}{ {sin}^{2} a + {cos}^{2}a - 2sin \: a \: cosa } } \\ \\ \\ \\ = > \sqrt{ \frac{1 + sin2a}{1 - sin2a} }$


Hence, Proved