Question

4 sin32°/cos58°+5 tan48°/cot42°-8 sec72°/cosec18°. without using trignometric table calculate

Answer 1

(4sin(90°-58°)/cos 58°)+(5tan(90°-42°)/cot42°)-(8sec(90°-18°)/cosec 18°)
=(4cos58°/cos 45°)+(5 cot 42°/cot 42°) -(8 cosec18°/cosec 18°)
=4+5-8
=9-8
=1

Answer 2

Hi ,

4( sin32/cos58)+5(tan48/cot42)-8(sec72/cosec18 )

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1 ) sin32/cos58

= sin( 90 - 58 )/cos58

= Cos 58 / cos 58

= 1 -----( a )

2 ) tan48/cot42

= Tan ( 90 - 42 )/ cot 42

= Cot 42 / cot 42

= 1 ------( b )

3 ) sec 72 / Cosec 18

= Sec ( 90 - 18 )/ Cosec 18

= Cosec 18 / Cosec 18

= 1 ------( c )
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According to the problem given ,

4 × 1 + 5 × 1 - 8 × 1 [ from ( a ) ,( b ),( c ) ]

= 4 + 5 - 8

= 9 - 8

= 1

I hope this helps you.

: )