Question

Cos70° + 4(sec59°-cot31°
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3sin20°

Sir ,on sec and cot there is square

Answer 1

Answer:

= Cos70/3Sin20 + 4($Sec^{2} 51 - Tan^{2}(90-31)/3-2/3Sin90$)

= Sin90-70/3Sin20 + 4($Sec^{2}59 - Tan^{2}59/3 - 2/3Sin90)$)

= 1/3 + 4(1)/3 - 2/3*1

= (1+4-2)/3

= 3/3

=1

Step-by-step explanation: