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If A = 60o, what is the value of : $\frac{[8cosA +7secA-tan^2A]}{10sin\frac{A}{2}}$ ? |
5 3 15 10 |
3 |
$\frac{[8cos60o +7sec60o- tan^260o]}{10sin\frac{60o}{2}}$ = $\frac{[8 × 1/2 + 7×2 - ( √3)2 ]}{10× 1/2 }$ = ( 4 + 14 - 3 ) ( 5) = 3 |
