Question

If SinA/SinB=m , and cosA/cosB=n , pove that: tanA=+- m/n1

Answer 1

Heya!!Here is your answer friend ⤵⤵
$\frac{sin \: \: a \: }{sin \: \: b} = m \: \: and \: \: \frac{cos \: \: a}{cos \: \: b \: } = n \\ \: \: \\ to \: prove = \: tan \: \: a \: \: \frac{ + }{ - } \frac{m}{n} \sqrt{ \frac{1 - {n}^{2} }{m {}^{2 } - 1} } \\ \\ sin \: \: b \: = \frac{sin \: \: a}{m} \: \: \: \: \: \: cos \: \: b = \frac{cos \: \: \: a \: }{n} \\ adding \: and \: squaring \: \: the \: equations \\ \\ sin {}^{2} \: a \: + \: \: cos {}^{2} \: \: b = \frac{sin {}^{2} a \: }{m {}^{2} } + \frac{cos {}^{2} a}{n {}^{2} } \\ dividing \: both \: sides \: by \: cos {}^{2} a \\ \\ sec {}^{2} a = \frac{1}{m {}^{2} } tan {}^{2} a \: \: + \frac{1}{n {}^{2} } \\ \\ 1 + tan {}^{2} a = \frac{1}{m {}^{2} } tan {}^{2} a + \frac{1}{n {}^{2} } \\ \\ tan {}^{2} a - \frac{1}{m {}^{2} } tan {}^{2} a \: = \frac{1}{n {}^{2} } - 1 \\ tan {}^{2} a(1 - \frac{1}{m {}^{2} } ) = \frac{1 - n {}^{2} }{n {}^{2} } \\ tan {}^{2} a( \frac{m {}^{2} - 1}{m {}^{2} } ) = \frac{1 - n {}^{2} }{n {}^{2} } \\ tan {}^{2} a = \frac{m {}^{2}(1 - n {}^{2} )}{n {}^{2} (m {}^{2} - 1)} \\ tan \: a \: = + - \frac{m}{n} \sqrt{ \frac{1 - n {}^{2} }{m {}^{2} - 1 } }$
Hence LHS= RHS