Question

sec 16 degrees cosec 74 degrees - cot 74 degrees tan 16 degrees​

Answer 1

$\sec(16) \csc(74) - \cot(74) \tan(16)$

$\sec(90 - 74) \csc(74) - \cot(74) \tan(90 - 74)$

$\csc(74) \csc(74) - \cot(74) \cot(74)$

${ \csc }^{2} 74 - { \cot }^{2} 74$

$1$

• Hence 1 is the answer.

here you should remember three formulae..

• $\sec(90 - \alpha ) = \csc( \alpha )$
• $\tan(90 - \alpha ) = \cot( \alpha )$
• ${ \csc}^{2} \alpha - { \cot}^{2} \alpha = 1$