Question

sec 48°÷cosec 42° + √3(tan 20°tan60°tan40°tan50°tan70°)
plzz help me

Answer 1

sec 48° / cosec 42° + √3 (tan 20° tan 70° tan 40° tan 50° tan 60° )
=cosec 42° / cosec 42° + √3(cot 70° tan 70° cot 50° tan 50° √3) [cosec (90°- A)=secA , cot(90°-A)=tanA and tan 60°=√3]
=1+√3(1/tan 70° *tan 70° 1/tan 50°* tan 50° √3) [cotA=1/tanA]
=1+√3(√3)
=1+3
=4

Answer 2

sec48/cosec(90-48)+√3[tan(90-70)√3tan(90-50)tan50tan70]
sec48/sec48+√3[cot70tan70√3cot50tan50]
1+√3[√3]
1+3
4