1.

Sin 72°sec 32°cos 189cosec 58°​

Answer»

Step-by-step EXPLANATION:

Given Question:-

The value of (sin72°/cos 18°)-(sec 32°/cosec58°)

Solution:-

Given that:

(sin72°/cos18°)-(sec32°/cosec58°)

=>[sin(90°-18°)/cos18°]-[sec(90°-58°)/cosec58°]

We know that sin(90°-A)=cos A

Sec (90°-A)=Cosec A

=>(Cos 18°/Cos 18°)-(Cosec 58°/Cosec 58°)

=>1-1

=>0

(or)

(sin72°/cos18°)-(sec32°/cosec58°)

=>[sin72°/cos(90°-72°)]-[sec32°/Cosec(90°-32°)]

We know that

Cos (90°-A)=Sin A

Cosec (90°-A)=Sec A

=>(sin 72°/sin72°)-(Sec 32°/sec32°)

=>1-1

=>0

Answer:-

The value of the given problem =0

Used formulae:-

  • Sin(90°-A)=Cos A
  • Cos (90°-A)=Sin A
  • Sec(90°-A)= Cosec A
  • Cosec (90°-A)= Sec A


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