Question

5cos²60°+ 4sec²30°-tan²45°÷ sin²30°+cos²30°​

Answer 1

5cos²60°+ 4sec²30°-tan²45°÷ sin²30°+cos²30°

= {5(1/2)^2 + 4(2/√3)^2} ÷ (1).

(we know , sin^2@+cos^2@ = 1 so,sin²30°+cos²30°=1)

= {5/4 + 14/3}

= (15+56)/12

= 71/12 (ans)

Answer 2

Answer:

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