Question

Cosec square 65- tan square 25/sin square 17 + sin square 73 plus one/under root 3 (tan 10 tan 30 tan 80

Answer 1

Cosec ²65-Tan²25/Sin²17+Sin²73+1/√3 *(tan10.tan30.tan80).
=sec²25-tan²25/sin²17+cos²17+1/√3 *(tan10.tan30.cot10).
=1+1/√3*tan30.
=1+1/3
=4/3

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)