Question

solve it ......10( i )

Answer 1

cosec62 = 1/sin62 but sin62 = sin(90 - 28) = cos28

⇒ cosec62 = 1/cos28

tan72 = tan(90 - 18) = cot18

tan54 = tan(90 - 36) = cot36

cot²(55) = cot²(90 - 35) = tan²35

we know that sec²θ - tan²θ = 1 and tanθ × cotθ = 1

⇒ (sec²35 - cot²55) = (sec²35 - tan²35) = 1

substituting all the above derived values in the question we get

= 2(1) - 1/tan30

= 2 - √3