Question

5cos^260°+4sec^230°-tan^245°/sin^230°+cos^230°​

Answer 1

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$\longrightarrow \tt \frac{5cos^260°+4sec^230°-tan^245°}{sin^230°+cos^230°} \\ \\ \longrightarrow \tt{\frac{5 { (\frac{1}{2}) }^{2} + 4 { (\frac{2}{ \sqrt{3} }) }^{2} - {(1)}^{2} }{ {( \frac{1}{2}) }^{2} + {( \frac{ \sqrt{3} }{2} )}^{2} } } \\ \\ \longrightarrow \tt{\frac{5( \frac{1}{4}) + ( \frac{16}{3}) - 1 }{ \frac{1}{4} + \frac{3}{4} } } \\ \\ \longrightarrow \tt{\frac{ \frac{15 + 64 - 12}{12} }{ \frac{4}{4} } } \\ \\ \longrightarrow \tt{\frac{67}{12}}$