Question

sec38/cosec52 +2√3 . tan17 tan38 tan60 tan52 tan73 – 3(sin²32 + sin²58). Evaluate it.

Answer 1

HELLO DEAR,

your questions is---------------> sec38/cosec52 + 2/√3*(tan17tan38 tan60tan52tan73) - 3(sin²32 + sin²58).

⇒sec(90 - 52)/cosec52 + 2/√3*{tan(90 - 73)tan73 tan(90 - 52) tan52 tan60} - 3{sin²(90 - 58) + sin²58}.


we know :-
sec(90 - Ф) = cosecФ
sin(90 - Ф) = cosФ
tan(90 - Ф) = cotФ
tan60 = √3
let us apply here,

⇒cosec52/cosec52 + 2/√3 * (cot73 *tan73* cot52*tan52*√3) - 3(cos²58 + sin²58)

[ sin²A + cos²A ]

⇒1 + 2(1/tan73 * tan73 * 1/tan52 * tan52) - 3(1)

⇒1 + 2 - 3

⇒3 - 3 = 0


I HOPE ITS HELP YOU DEAR,
THANKS