Question

please tell me 1) if we can add degrees in the second question by taking tan5 common ?
2) the explanation of both question s

Answer 1

1) tan∅ = 4


{ tan∅}/{ sin³∅/cos∅ + sin∅.cos∅}

={ tan∅cos∅ }/{ sin³∅ + cos²∅.sin∅}

=( tan∅.cos∅)/{ sin∅( sin²∅ +cos²∅)}

=(tan∅.cos∅)/(sin∅.1)

[ sin²∅ + cos²∅ =1 use this


= tan∅/(sin∅/cos∅)

=tan∅/tan∅

= 1
here option is wrong

2) tan5°.tan10°.tan15°.....tan85°

we know,
tan5° = tan(90-5) =cot85°
tan5.tan85° = 1
similarly ,

tan10.tan80° = 1
tan15°.tan75° = 1
..........
....
tan40.tan50° = 1

so,
(tan5°.tan85°).(tan10°.tan80°).(tan15°.tan75°).....(tan40°.tan50°).tan45°

= 1 × 1 × 1 .....× 1= 1

Answer 2

Answer:

1) tan 0 = 4

{ tan{ sin°Ø/cos Ø + siØ.cos Ø}

=( tabaco }{ sin + cos?Ø.sinØ}

=( tano.cos)/{ sin Øl sin?Ø +cos²Ø}

=(tanØ.cos)/(sin Ø.1)

[ sin?Ø + cos2Ø =1 use this

= tanØ/(sinØ/cosØ)

=tan /tano

= 1

here option is wrong

2) tan 5°.tan 1 O°.tan15°..tan85°

we know,

tan5° = tan(90-5) =cot85°

tan5.tan85° = 1 similarly,

tan10.tan80° = 1 tan15°.tan75º = 1

tan40.tan50° = 1

so,

(tan5°.tan85°).(tan10°.tan80°). (tan15°.tan75).(tan40°.tan50°).tan45° ...

= 1x 1 x 1 ..x 1= 1