Question

(sina+coseca) ²+(cosa+seca) ²=7+tan²a+cot²a​

Answer 1

TO PROVE :-

$\\ \sf \: (sinA + {cosecA)}^{2} + (cosA + {secA)}^{2} = 7 + {tan}^{2}A + {cot}^{2}A$

SOLUTION :-

$\\ \sf \: LHS = (sinA + {cosecA)}^{2} + (cosA + {secA)}^{2} \\ \\ \boxed{ \bf \:cosecA = \dfrac{1}{sinA} } \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \:secA = \dfrac{1}{cosA} } \\ \\ \implies \sf \: { \left(sinA + \dfrac{1}{sinA} \right)}^{2} + {\left( cosA + \dfrac{1}{cosA} \right)}^{2}$

$\\ \implies \sf \: {sin}^{2}A + {\left( \dfrac{1}{sinA} \right)}^{2} + 2( \cancel{sinA})\left( \dfrac{1}{ \cancel{sinA}} \right) + {cos}^{2}A + {\left( \dfrac{1}{cosA} \right)}^{2} + 2( \cancel{cosA})\left( \dfrac{1}{ \cancel{cosA}} \right)$

$\\ \implies \sf \: {sin}^{2}A + \dfrac{1}{ {sin}^{2}A } + 2 + {cos}^{2}A + \dfrac{1}{ {cos}^{2}A } + 2 \\ \\ \implies \sf \: {sin}^{2}A + {cos}^{2}A + \dfrac{1}{ {sin}^{2}A } + \dfrac{1}{ {cos}^{2}A } + 4 \\ \\ \boxed{ \bf \: {sin}^{2}A + {cos}^{2}A = 1 }$

$\\ \implies \sf \: 1 + \dfrac{ {sin}^{2}A + {cos}^{2}A }{ {sin}^{2}A. {cos}^{2}A } + 4 \\ \\ \implies \sf \: 5 + \dfrac{1}{ {sin}^{2}A. {cos}^{2}A } \\ \\ \boxed{ \bf \:sinA = \dfrac{1}{cosecA} } \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \:cosA = \dfrac{1}{secA} } \\ \\ \implies \sf \: 5 + ( {cosec}^{2}A) ( {sec}^{2}A )$

$\\ \boxed{ \bf \: {cosec}^{2}A = 1 + {cot}^{2}A } \: \: \: \: \: \: \boxed{ \bf \: {sec}^{2}A = 1 + {tan}^{2}A } \\ \\ \implies \sf \: 5 + (1 + {cot}^{2}A )(1 + {tan}^{2}A ) \\ \\ \implies \sf \: 5 + 1 + {tan}^{2}A + {cot}^{2}A + ( {cot}^{2}A . {tan}^{2}A ) \\ \\ \boxed{ \bf \: cotA = \dfrac{1}{tanA} }$

$\\ \implies \sf \: 6 + {tan}^{2}A + {cot}^{2}A + \left( \dfrac{1}{ \cancel{ {tan}^{2}A }} \right)( \cancel{{tan}^{2}A }) \\ \\ \implies \sf \: 7 + {tan}^{2}A + {cot}^{2}A = RHS \: \: \: \: (verified)$

Answer 2

Question :

To Prove :

$\\ \sf \: (sinA + {cosecA)}^{2} + (cosA + {secA)}^{2} = 7 + {tan}^{2}A + {cot}^{2}A$

Solution :

$\\ \sf \: LHS = (sinA + {cosecA)}^{2} + (cosA + {secA)}^{2} \\ \\ \blue{ \boxed{ \bf \:sinA = \dfrac{1}{cosecA} } \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \:cosA = \dfrac{1}{secA} }} \\ \\ \implies \sf \: { \left(sinA + \dfrac{1}{sinA} \right)}^{2} + {\left( cosA + \dfrac{1}{cosA} \right)}^{2}$

$\\ \implies \sf \: {sin}^{2}A + {\left( \dfrac{1}{sinA} \right)}^{2} + 2( \cancel{sinA})\left( \dfrac{1}{ \cancel{sinA}} \right) + {cos}^{2}A + {\left( \dfrac{1}{cosA} \right)}^{2} + 2( \cancel{cosA})\left( \dfrac{1}{ \cancel{cosA}} \right)$

$\\ \implies \sf \: {sin}^{2}A + \dfrac{1}{ {sin}^{2}A } + 2 + {cos}^{2}A + \dfrac{1}{ {cos}^{2}A } + 2 \\ \\ \implies \sf \: {sin}^{2}A + {cos}^{2}A + \dfrac{1}{ {sin}^{2}A } + \dfrac{1}{ {cos}^{2}A } + 4 \\ \\ \blue{\boxed{ \bf \: {sin}^{2}A + {cos}^{2}A = 1 } }$

$\\ \implies \sf \: 1 + \dfrac{ {sin}^{2}A + {cos}^{2}A }{ {sin}^{2}A. {cos}^{2}A } + 4 \\ \\ \implies \sf \: 5 + \dfrac{1}{ {sin}^{2}A. {cos}^{2}A } \\ \\\blue{ \boxed{ \bf \:sinA = \dfrac{1}{cosecA} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{\boxed{ \bf \:cosA = \dfrac{1}{secA} }} \\ \\ \implies \sf \: 5 + ( {cosec}^{2}A) ( {sec}^{2}A )$

$\\ \blue{\boxed{ \bf \: {cosec}^{2}A = 1 + {cot}^{2}A }} \: \: \: \: \: \: \blue{\boxed{ \bf \: {sec}^{2}A = 1 + {tan}^{2}A }} \\ \\ \implies \sf \: 5 + (1 + {cot}^{2}A )(1 + {tan}^{2}A ) \\ \\ \implies \sf \: 5 + 1 + {tan}^{2}A + {cot}^{2}A + ( {cot}^{2}A . {tan}^{2}A ) \\ \\ \blue{\boxed{ \bf \: tanA = \dfrac{1}{cotA} }}$

$\\ \implies \sf \: 6 + {tan}^{2}A + {cot}^{2}A + \left( \dfrac{1}{ \cancel{ {tan}^{2}A }} \right)( \cancel{{tan}^{2}A }) \\ \\ \implies \sf \: 7 + {tan}^{2}A + {cot}^{2}A$

R.HS.

$\huge \underline{Hence\: \: Proved }$

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