Question

Cos43=m2-1/m2+1,then sin43+cos47=

Answer 1

Answer:

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Step-by-step explanation:

$To \: find = \\ value \: of \: sin \: 43 + cos \: 47 \: (in \: terms \: of \: m) \\ \\ so \: here \: cos \: 43 = \frac{m {}^{2} - 1 }{m {}^{2} + 1 } \\ \\ so \: we \: know \: that \: \\ sin {}^{2} θ + cos { }^{2} θ = 1 \\ where \: θ = 43 \\ \\ hence \: accordingly \\ sin {}^{2} \: 43 = 1 - cos {}^{2} .43 \\ = 1 - ( \frac{m {}^{2} - 1}{m {}^{2} + 1} ) {}^{2} \\ \\ = 1 - \frac{m {}^{4} + 1 - 2m {}^{2} }{m {}^{4} + 1 + 2m {}^{2} } \\ \\ = \frac{m {}^{4} + 1 + 2m {}^{2} - (m {}^{4} + 1 - 2m {}^{2} )}{m {}^{4} + 1 + 2m {}^{2} } \\ \\ = \frac{m {}^{4} + 1 + 2m {}^{2} - m {}^{4} - 1 + 2m {}^{2} }{m {}^{4} + 1 + 2m {}^{2} } \\ \\ = \frac{2m {}^{2} + 2m {}^{2} }{(m {}^{2} + 1) {}^{2} } \\ \\ = \frac{4m {}^{2} }{(m {}^{2} + 1) {}^{2} } \\ \\ = \frac{(2m) {}^{2} }{(m {}^{2} + 1) {}^{2} } \\ \\ = ( \frac{2m}{m {}^{2} + 1} ) {}^{2} \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ \\ sin \: 43 = \frac{2m}{m {}^{2} + 1 }$

$now \: similarly \\ we \: know \: that \\ according \: to \: complementary \: angle \: relation \\ sin \: (90 - θ) = cos \: θ \\ so \: hence \: \\ sin \: 43 = cos \: (90 - 43) \\ = cos \: 47 \\ \\ hence \: cos \: 47 = \frac{2m}{m {}^{2} + 1 }$

$so \: hence \\ sin \: 43 + cos \: 47 \\ = \frac{2m}{m {}^{2} + 1} \: + \: \frac{2m}{m {}^{2} + 1} \\ \\ = \frac{2m + 2m}{m {}^{2} + 1} \\ \\ = \frac{4m}{m {}^{2} + 1}$