1.

Prove that `cosec^(2)22^(@)cot^(2)68^(@)=sin^(2)22^(@)+sin^(2)68^(@)+cot^(2)68^(@)`.

Answer» `cosec^(2)22^(@)cot^(2)68^(@)=cosec^(2)(90^(@)-68^(@))cot^(2)68^(@)`
`=sec^(2)68^(@)cot^(2)68^(@)=(1+tan^(2)68^(@))cot^(2)68^(@)`
`=cot^(2)68^(@)+tan^(2)68^(@)cot^(2)68^(@)`
`=cot^(2)68^(@)+tan^(2)68^(@)cot^(2)68^(@)`
`=cot^(2)68^(@)+tan^(2)68^(@).(1)/(tan^(2)68^(@))`
`=cot^(2)68^(@)+1`
`=cot^(2)68^(@)+sin^(2)22+cos^(2)(90^(@)-68^(@))`
`=cot^(2)68^(@)+sin^(2)22+sin^(2)68^(@)`
`=sin^(2)22^(@)+sin^(2)68^(@)+cot^(2)68^(@)`.
Hence `cosec^(2)22^(@)cot^(2)68^(@)=sin^(2)22^(@)+sin^(2)68^(@)+cot^(2)68^(@)`


Discussion

No Comment Found