Question

sin72°÷cos18° - sec32°÷cosec58°

Answer 1

sin 72 /cos 18 - sec 32/cosec 58

sin 72 = cos (90-72) and sec 32 = cosec (90-32)

= cos (90-72)/cos (18) - cosec (90-32)/cosec (58)

cos (18)/cos (18) - cosec (58)/cosec (58)

Therefore

1 - 1 = 0

0

Hope it helps

Answer 2

Hey frnd here is ur ans.....

sin72°/cos18° - sec32°/cosec58°

sin72°/cos(90°-72°) - sec32°/cosec(90°-32°)

sin72°/sin72° - sec32°/sec32°

1-1 =0

Ans... ;)

Hope it helps you✌✌

Thankyou ☺☺

sweety ⚡⚡