Question

sec A+ tan A = 2 then sec A is equal to

Answer 1

Hope it helps

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Answer 2

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



$\textsf{ Using trigonometric identity : }$


$\boxed{\mathsf{ \implies tan \: A \: = \: \sqrt{ \: {sec}^{2} \: A \: - \: 1 \: \: }} }$


$\mathsf{\implies sec\: A \: + \: tan \: A \: = \: 2 } \\ \\ \mathsf{\implies sec\: A \: + \: \sqrt{\: {sec}^{2} \: A \: - \: 1 \: \: } \: = \: 2 }$

$\\ \mathsf{ \implies \sqrt{\: {sec}^{2} \: A \: - \: 1 \: \:} \: = \: 2 \: - \: sec \:A } \\ \\ \mathsf{\implies {sec}^{2} \: A \: - \: 1 \: = \:{ ( 2 \: - \: sec \: A )}^{2}} \\ \\ \textsf{ Using Algebraic identity : } \\ \\ \boxed{\mathsf{ \implies {( a \: - \: b )}^{2} \: = \: {a}^{2} \: + \: {b}^{2} \: - \: 2ab }}$

$\mathsf{ \implies {sec}^{2} \: A \: - \: 1 \: = \: {(2)}^{2} \: + \: { ( sec \: A )}^{2} \: - \: 2 \: \cdot \: \: 2 \: \cdot \: sec \: A} \: \\ \\ \mathsf{ \implies \cancel{{sec}^{2} \: A} \: - \: 1 \: = \: 4 \: + \: \: \cancel{{sec }^{2} \: A }\: - \: 4 \: sec \: A } \\ \\ \mathsf{ \implies - 1 \: - \: 4 \: = \: - 4 \: sec \:A \: } \\ \\ \mathsf{ \implies - 5 \: = \: - 4 \: sec \:A \: } \\ \\ \mathsf{ \implies sec \: A \: = \: \dfrac{ \cancel{- } \: 5}{ \cancel{ -} \: 4 \: } } \\ \\ \mathsf{ \therefore \: \: sec \: A \: = \: \frac{5}{4} }$

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