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माना ACB एक समकोण त्रिभुज इस प्रकार है कि `angle C=90^(@), AB=29` इकाई `BC=21` इकाई तथा `angle ABC=theta,` तब निम्न के मान ज्ञात कीजिए-
`(i) cos ^(2)theta+sin^(2)" "(ii) cos ^(2)theta-sin ^(2) theta`

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`DeltaABC` में पाइथागोरस प्रमेय का प्रयोग करने पर,
`AB^(2)=AC^(2)+BC^(2)`
`impliesAC=sqrt(AB^(2)-BC^(2))=sqrt(29^(2)-21^(2))`
`=sqrt(400)=20` इकाई
इसलिए `sin theta=("लम्ब")/(" कर्ण")=(AC)/(AB)=20/29`
तथा `cos theta =(" आधार")/(" कर्ण")=(BC)/(AB)=21/29`
अब
(i) ` cos ^(2)theta+sin ^(2)theta=((21)/(29))^(2)+((20)/(29))=441/841+400/841=(441+400)/(841)=841/841=1`
(ii)`cos ^(2)theta-sin^(2)theta=((21)/(29))^(2)-((20)/(29))^(2)=441/841-400/841=41/841`
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