Question

if secA - tanA = x, show that secA + tanA = 1/x and hence find the values of cosA and sinA

Answer 1

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



$\underline{\textsf{Given,}} \\ \\ \mathsf{\implies sec \: A \: - \: tan \: A \: = \: x \quad...(1) } \\ \\ \underline{\textsf{To Prove :}} \\ \\ \mathsf{\implies sec \: A \: + \: tan \: A \: = \: \dfrac{1}{x}}$

$\textsf{By Trigonometric Identity , } \\ \\ \mathsf{\implies sec^2 \: A \: - \: tan^2 \: A \: = \: 1 } \\ \\ \mathsf{\implies ( sec \: A \: + \: tan \: A ) ( sec \: A \: - \: tan \: A ) \: = \: 1 } \\ \\ \textsf{Plug the value of ( 1 ),} \\ \\ \mathsf{\implies ( sec \: A \: + \: tan \: A)x \: = \: 1 } \\ \\ \mathsf{\therefore \: \: sec \: A \: + \: tan \: A \: = \: \dfrac{1}{x}} \\ \\ \underline{\underline{ \textsf{Proved !!}}}$



$\underline{\textsf{Now,}} \\ \\ \mathsf{\implies sec \: A \: - \: tan \: A \: = \: x \quad...(1)} \\ \\ \mathsf{\implies sec \: A \: + \: tan \: A \: = \: \dfrac{1}{x} \quad...(2)}$


$\textsf{Add both the equations , } \\ \\ \mathsf{ \implies sec \: A \: - \: \cancel{tan \: A} \: + \: sec \: A \: + \: \cancel{ tan \: A} \: = \: x \: + \: \dfrac{1}{x} }$



$\mathsf{\implies 2 sec \: A = \: \dfrac{x^2 \: + \: 1}{x}} \\ \\ \mathsf{\implies sec \: A = \: \dfrac{x^2 \: + \: 1}{2x}} \\ \\ \mathsf{\implies \dfrac{1}{cos \: A } \: = \: \dfrac{x^2 \: + \: 1 }{2x}} \quad \mathsf{ \left\{ cos \: \theta \: = \: \dfrac{1}{sec \: \theta} \right\}} \\ \\ \mathsf{\therefore \: cos \: A \: = \: \dfrac{2x}{x^2 \: + \: 1 }}$


$\textsf{Using Trigonometric Identity :} \\ \\ \mathsf{\implies sin \: A \: = \: \sqrt{1 \: - \: cos^2 A}}$


$\mathsf{\implies sin \: A \: = \: \sqrt{1 \: - \: \left( \dfrac{2x}{x^2 \: + \: 1} \right)^2 }} \\ \\ \mathsf{\implies sin \: A \: = \: \sqrt{1 \: - \: \dfrac{4 {x}^{2} }{ {( {x}^{2} \: + \: 1) }^{2} } } }$


$\mathsf{\implies sin \: A \: = \: \sqrt{ \dfrac{( {x}^{2} \: + \: 1) {}^{2} \: - \: 4 {x}^{2} }{( {x}^{2} \: + \: 1)^{2} } }} \\ \\ \mathsf{\implies sin \: A \: = \: \sqrt{ \dfrac{ {x}^{4} \: + \: 1 \: + \: 2 {x}^{2} \: - \: 4 {x}^{2} }{( {x}^{2} \: + \: 1 )^{2} } }}$


$\mathsf{\implies sin \: A \: = \: \sqrt{ \dfrac{ {x}^{4} \: + \: 1 \: - \: 2 {x}^{2} }{( {x}^{2} \: + \: 1) ^{2} } }} \\ \\ \mathsf{\implies sin \: A \: = \: \sqrt{ \dfrac{( {x}^{2} \: - \: 1) {}^{2} }{( {x}^{2} \: + \: 1 ) ^{2} } }} \\ \\ \mathsf{\implies sin \: A \: = \: \sqrt{ \left( \frac{ {x}^{2} \: - \: 1 }{ {x}^{2} \: + \: 1 } \right) ^{2} }}$


$\mathsf{ \: \therefore sin \: A \: = \: \dfrac{ {x}^{2} \: - \: 1 }{ {x}^{2} \: + \: 1 } }$

Answer 2

This may be correct...