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Prove that (sin theta + Cosec theta)^2+(cos theta +sec theta)^2 = 7+ tan^2 theta + cot^2 theta |
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Answer» LHS = (sin^2 theta + cosec^2 theta + 2×sin theta × cosec theta) + ( cos^2 theta + sec^2 theta + 2×cos theta × sec theta) => (sin^2 theta + cos^2 theta)+ cosec^2 theta + sec^2 theta + 2×sin theta×cosec theta + 2 ×cos theta × sec theta => 1 + (1+cot^2 theta) + (1+tan^theta) + (2×sin theta× 1/sin theta ) + ( 2× cos theta × 1/cos theta) => 7 + cot^2 theta + tan^2 theta... Now, LHS = RHS So, it\'s prove.... Wlcm..?muh kyu bana rkha h Thank you ? ?? Cosec ko sin.....and sec ko cos me change kar ke......do as given ...i mean perform their square....you definately get it,???? Used identity=> sin theta cosec theta=1, cos theta sec theta=1, sin^2theta+cos^2theta=1, cosec^2theta=1+cot^2theta. LHS= (sin theta+cosec theta)^2+(cos thetha+ sec thetha)^2(sin^2theta+coscec^2theta+2 sin theta cosec theta) + (cos^2theta+ sec^2theta+2cos theta sec theta) (sin^2theta+cosec^2theta+2) + (cos^2theta+sec^2theta+2) (sin^2theta+cos^2theta)+4+(cosec^2 thetha+sec^2theta) 1+4+(1+cot^2theta)+(1+tan^2 theta) = (7+tan^2theta+cot^2) Anybody here solve this ?? |
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