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