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What is the value of $\frac{3cosec42^o}{sec48^o}-\frac{5cos32^o}{sin58^o}$ |
-1 5 0 -2 |
-2 |
\(\frac{3 cosec42º}{sec48º}\) - \(\frac{5 sec32º}{cosec58º}\) { If A + B = 90º , Then secA = cosecB } = \(\frac{3 sec48º}{sec48º}\) - \(\frac{5 sec32º}{sec32º}\) = 3 - 5 = -2
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