What is the value of $\frac{3cosec42^o}{sec48^o}-\frac{5cos32^o}{sin58^o}$
Answer & explanation
Correct answer: option 4
\(\frac{3 cosec42º}{sec48º}\) - \(\frac{5 sec32º}{cosec58º}\)
{ If A + B = 90º , Then secA = cosecB }
= \(\frac{3 sec48º}{sec48º}\) - \(\frac{5 sec32º}{sec32º}\)
= 3 - 5
= -2