Question

Prove that
secA- tanA = 1- tan(A/2)/1+tan (A/2)​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: secA - tanA$

$\rm \: =  \: \dfrac{1}{cosA} - \dfrac{sinA}{cosA}$

$\rm \: =  \: \dfrac{1 - sinA}{cosA}$

can be further rewritten as

$\rm \: =  \: \dfrac{1 - cos\bigg(\dfrac{\pi}{2} - A\bigg) }{sin\bigg(\dfrac{\pi}{2} - A\bigg)}$

Let assume that

$\boxed{\sf{  \: \: \: \dfrac{\pi}{2} - A \: = \: x \: \: }}$

So, above expression can be rewritten as

$\rm \: =  \: \dfrac{1 - cosx}{sinx}$

$\rm \: =  \: \dfrac{2 {sin}^{2}\dfrac{x}{2} }{2 \: sin\dfrac{x}{2} \: cos\dfrac{x}{2} }$

$\rm \: =  \: tan\dfrac{x}{2}$

On substituting the value of x, we get

$\rm \: =  \: tan\bigg(\dfrac{\pi}{4} - \dfrac{A}{2} \bigg)$

$\rm \: =  \: \dfrac{tan\dfrac{\pi}{4} - tan\dfrac{A}{2} }{1 + tan\dfrac{\pi}{4} \: \: tan\dfrac{A}{2}}$

$\rm \: =  \: \dfrac{1 - tan\dfrac{A}{2} }{1 + 1 \times \: \: tan\dfrac{A}{2}}$

$\rm \: =  \: \dfrac{1 - tan\dfrac{A}{2} }{1 + \: tan\dfrac{A}{2}}$

Hence,

$\\ \rm\implies \:\boxed{\sf{  \:\rm \: secA - tanA=  \: \dfrac{1 - tan\dfrac{A}{2} }{1 + \: tan\dfrac{A}{2}} \: }}$

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Formulae Used :-

$\boxed{\sf{  \: \: sin2A = 2sinAcosA \: \: }}$

$\boxed{\sf{  \: \: cos2A = 1 - {2sin}^{2}A \: \: }}$

$\boxed{\sf{  \: \: tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} \: }}$

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Additional Information

$\boxed{\sf{  \: \: cos2A = {2cos}^{2}A - 1 = {cos}^{2}A - {sin}^{2}A \: }}$

$\boxed{\sf{  \: \: sin2A \: = \: \frac{2tanA}{1 + {tan}^{2} A} \: \: }}$

$\boxed{\sf{  \: \: cos2A \: = \: \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A} \: \: }}$

$\boxed{\sf{  \: \: tan2A \: = \: \frac{2tanA}{1 - {tan}^{2} A} \: \: }}$

$\boxed{\sf{  \: \: sin3A = 3sinA \: - \: {4sin}^{3}A \: \: }}$

$\boxed{\sf{  \: \: cos3A = {4cos}^{3}A - 3cosA \: \: }}$

$\boxed{\sf{  \: \: tan3A = \frac{3tanA - {tan}^{3}A }{1 - {3tan}^{2} A} \: }}$

Answer 2

$⟹ secA−tanA=$

$−tan2A$

$1+tan2 A1$