Question

sec 37/cosec 53+2 cot 15 cot 25 cot 45 cot 7 5 cot 65

Answer 1

$\frac{ \sec(37) }{ \csc(53) } + 2 \cot(15) \cot(25) \cot(45) \cot(75) \cot(65) \\ = \frac{ \sec(37) }{ \sec(37) } + 2 \cot(15) \cot(25) \cot(45) \tan(15) \tan(25) \\ = 1 + 2 \cot(15) \cot(25) \times 1 \times \frac{1}{ \cot(15) } \times \frac{1}{ \cot(25) } \\ = 1 + 2 \\ = 3$

Answer 2

Sec (90-53)/ Cosec 53 + 2 Cot (90-75) Cot (90-65).1.Cot 75 Cot 65

Cosec 53/Cosec 53  +2 tan 75. tan 65. 1. 1/tan 75 . 1/ tan 65 [ cot 45=1,                                                             sec (90-A) = Cosec A, Cot(90-A) = tan A]

after cancellation
1+2
=3