Question

Without using trigonometric table ,evaluate
cosec2 65°- tan2 25°/sin2 17° + sin2 73° + 1/√3(tan 10° × tan 30° × tan 80°)

Plz help me..........

Answer 1

refer to the attachment

Answer 2

Thank you for asking this question. Here is your answer:

csc² 65 = csc²(90-25) = sec² 25

So the part csc² 65 - tan² 25 = 1

tan 80 = tan (90-10) = cot 10 = 1/tan 10 -- (1)

So tan 10 x tan 80 = 1 from (1)

Sin 17 = sin(90-73) = cos(73) --- (2)

So sin² 17 + sin² 73 = 1  from (2)

1 / (1 + 1/√3 x 1 x tan 30) = 1 / (1 + 1/3)

= 1/(4/3)

= 3/4

(since tan 30 = 1/√3)