Question

evaluate without using trigonometric table

Answer 1

Hello !!

HERE IS YOUR ANSWER,,

$2( { \sec}^{2}35 - { \cot }^{2} 55) - \frac{ \cos(28) \times \csc(62) }{ \tan(18) \times \tan(36) \times \tan(54) \times \tan(72) } \\ \\ \\ 2( { \sec}^{2} 35 - { \tan }^{2} (90 - theta)) - \frac{ \sin(90 - 28) \times \csc(62) }{ \cot(90 - 18) \times \tan(72) \times \cot(90 - 54) \times \tan(36) } \\ \\ \\ 2( { \sec }^{2} 35 - { \tan}^{2} 35) - \frac{ \sin(62) \times \csc(62) }{ \cot(72) \times \tan(72) \times \cot(36) \times \tan(36) } \\ \\ 2(1) - \frac{1}{1 \times 1} \\ \\ 2 - 1 \\ \\ 1 \\ \\ \\ \\ hope \: this \: will \: help \: you \: \\ \\ \\ thanks$
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