Question

If a*secβ+b*tanβ=1 and a²sec²β - b²tan²β=5, then a²b² + 4a² is ?

Answer 1

$\underline{\mathfrak{Solution : }}$

$\text{Given, }$

$\mathsf{ \implies a\: sec\: \beta \: + \: b \: tan \: \beta \: = \: 1 \qquad...(1) }$

$\mathsf{ \implies {a}^{2} {sec}^{2} \beta \: - \: {b}^{2} {tan}^{2} \beta \: = \: 5 \qquad...(2)}$

$\mathsf{ \implies (a \: sec \: \beta)^{2} \: - \: (b \: tan \: \beta)^{2} \: = \: 5}$



$\text{Using Algebraic identity : } \\ \\ \\ \boxed{\mathsf{\implies {a}^{2} \: - \: {b}^{2} \: = \: ( a \: + \: b )(a \: - \: b )}}$

$\mathsf{ \implies (a \: sec \: \beta \: + \: b \: tan \: \beta)(a \: sec \: \beta \: - \: b \: tan \: \beta ) \: = \: 5} \\ \\ \\ \text{ Substitute the value of eq.n ( 1 ), } \\ \\ \\ \mathsf{ \implies 1(a \: sec \: \beta \: - \: b \: tan \: \beta) \: = \: 5} \\ \\ \\ \mathsf{ \therefore \: \: a \: sec \: \beta \: - \: b \: tan \: \beta \: = \: 5 \qquad...(3)}$

$\text{Adding eq.n(1) and eq.(3)} \\ \\ \\ \mathsf{ \implies a \: sec \: \beta \: + \: \cancel{b \: tan \: \beta }\: + \: a \: sec \: \beta \: - \cancel{\: b \: tan \: \beta} \: = \: 1 \: + \: 5} \\ \\ \\ \mathsf{ \implies 2 a\: sec \: \beta \: = \: 6} \\ \\ \\ \mathsf{ \implies a \: = \: \dfrac{6}{2 \: sec \: \beta}} \\ \\ \\ \mathsf{ \implies a \: = \: \dfrac{3}{sec \: \beta}}$

$\text{Substitute the value of \textbf{a} in eq.n (1),}$

$\mathsf{ \implies a \: sec \: \beta \: + \: b \: tan \: \beta \: = \: 1} \\ \\ \\ \mathsf{ \implies \dfrac{3}{ \cancel{sec \: \beta}} \: \times \: \cancel{sec \: \beta} \: + \: b \: tan \: \beta \: = \: 1} \\ \\ \\ \mathsf{ \implies 3 \: + \: b \: tan \: \beta \: = \: 1} \\ \\ \\ \mathsf{ \implies b \: tan \: \beta \: = \: 1 \: - \: 3} \\ \\ \\ \mathsf{ \implies b \: tan \: \beta \: = \: - 2} \\ \\ \\ \mathsf{ \implies b \: = \: \frac{ - 2}{tan \: \beta} }$

$\text{Now, }$

$\mathsf{ = {a}^{2} {b}^{2} \: + \: 4 {a}^{2} } \\ \\ \\ \mathsf{ = (ab)^{2} \: + \: (2a)^{2} }$

$\text{Plug the value of \textbf{a} and \textbf{b},} \\ \\ \\ \mathsf{ = { (\dfrac{3}{sec \: \beta } \times \dfrac{ - 2}{tan \: \beta} )}^{2} \: + \: {(2 \: \times \: \dfrac{3}{sec \: \beta} )}^{2} }$

$\text{Using trigonometric identity : }\\ \\ \\ \boxed{\mathsf{\implies sec \: \theta \: = \: \dfrac{1}{cos \: \theta} , \implies tan\: \theta \: = \: \dfrac{sin \: \theta}{cos \: \theta}}}$

$\mathsf{ = ( { \dfrac{ \: \: \: \: \: \: \: \: \: 3 \: \: \: \: \: \: \: }{ \dfrac{1}{cos \: \beta} } \times \dfrac{ \qquad - 2 \qquad}{ \dfrac{sin \: \beta}{cos \: \beta}} })^{2} \: + \: ( \dfrac{ \qquad6 \qquad}{ \dfrac{1}{cos \: \beta}})^{2} }$

$\mathsf{ =(3 \: cos \: \beta \: \times \: \dfrac{ - 2 \: cos \: \beta}{sin \: \beta})^{2} \: + \: (6 \: cos \: \beta) ^{2} } \\ \\ \\ \mathsf{ = ( \dfrac{ - 6 \: {cos}^{2} \: \beta }{sin \: \beta} )^{2} \: + \: 36 \: {cos}^{2} \: \beta }$

$\mathsf{ = \dfrac{36 \: {cos}^{4} \beta}{sin^{2} \: \beta } \: + \: 36 \: {cos}^{2} \beta }$

$\mathsf{ = \dfrac{36 \: {cos}^{4} \beta \: + \: 36 \: {cos}^{2} \beta \: \times \: {sin}^{2} \beta }{ {sin}^{2} \: \beta }}$

$\mathsf{ = \dfrac{36 \: {cos}^{2} \beta( {cos}^{2} \beta \: + \: {sin}^{2} \beta) }{sin^{2} \beta}}$

$\text{Using trigonometric identity,} \\ \\ \boxed{\mathsf{\implies ( {sin}^{2} \theta \: + \: {cos}^{2}\theta ) \: = \: 1 }}$

$\mathsf{ \implies \dfrac{36 \: cos^{2} \beta(1)}{ {sin}^{2} \beta }} \\ \\ \\ \mathsf{ = \dfrac{36 \: {cos}^{ 2 } \beta }{ {sin}^{2} \beta } } \\ \\ \\ \mathsf{ = 36( { \dfrac{cos \: \beta}{sin \: \beta} )}^{2} }$

$\text{Using trigonometric identity,} \\ \\ \\ \boxed{\mathsf{\implies \dfrac{cos \: \theta}{sin \: \theta } \: = \: cot \: \theta}}$

$\mathsf{ = \: 36 \: {(cot \: \beta)}^{2} } \\ \\ \\ \mathsf{ = 36 \: {cot}^{2} \beta}$

$\boxed{\underline{ \: \mathfrak{Hope \: \: it \: \: helps \: \: !! } \: }}$