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tan A ÷ ( 1 + tan2 A )2 + cot A ÷ ( 1 + cot2\xa0A )2\xa0= sin A cos A |
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Answer» Take LHS{tex}{tan A\\over (1+tan^2A)^2}+{cotA\\over (1+cot^2A)^2}{/tex}{tex}= {tan A\\over sec^4A}+{cotA\\over cosec^4A}{/tex}{tex}={sin A\\over cosAsec^4A}+{cosA\\over sinAcosec^4A}{/tex}{tex}= {sin A\\over sec^3A}+{cosA\\over cosec^3A}{/tex}{tex}= {sin Acos^3A}+{cosAsin^3A}{/tex}{tex}= {sin AcosA}({cos^2A+sin^2A}){/tex}=> sinAcosA = RHSHence Verified\xa0 {tex}{tan A\\over (1+tan^2A)^2}+{cotA\\over (1+cot^2A)^2}= sinAcosA{/tex}Taking LHS{tex}{tan A\\over (1+tan^2A)^2}+{cotA\\over (1+cot^2A)^2}{/tex}=\xa0{tex}{tan A\\over (sec^2A)^2}+{cotA\\over (cosec^2A)^2}{/tex}=\xa0{tex}{tan A\\over sec^4A}+{cotA\\over cosec^4A}{/tex}=\xa0{tex}{sin A\\over cosAsec^4A}+{cosA\\over sinAcosec^4A}{/tex}{tex}= {sin A\\over sec^3A}+{cosA\\over cosec^3A}{/tex}\xa0{tex}= {sin Acos^3A}+{cosAsin^3A}{/tex}{tex}= {sin AcosA}({cos^2A+sin^2A}){/tex}= sinAcosA = RHSHence proved\xa0 |
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