Question

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1/secA+tanA - 1/cosA=1/cosA-1/secA-tanA
Prove it​

Answer 1

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

Consider, LHS

$\rm :\longmapsto\:\dfrac{1}{secA + tanA} - \dfrac{1}{cosA}$

On rationalizing the first term,

$\rm \:  =  \: \dfrac{1}{secA + tanA} \times \dfrac{secA - tanA}{secA - tanA} - secA$

We know,

$\boxed{ \bf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2} }}$

So, using this, we get

$\rm \:  =  \: \dfrac{secA - tanA}{ {sec}^{2} A - {tan}^{2} A} - secA$

We know,

$\boxed{ \bf{ \: {sec}^{2}x - {tan}^{2}x = 1}}$

So, using this, we get

$\rm \:  =  \: secA - tanA - secA$

$\rm \:  =  \: - \: tanA$

So,

$\rm :\longmapsto\:\boxed{ \bf{ \:\dfrac{1}{secA + tanA} - \dfrac{1}{cosA} = - \: tanA}} - - (1)$

Now, Consider,

$\rm :\longmapsto\:\dfrac{1}{cosA} - \dfrac{1}{secA + tanA}$

On rationalizing the second term, we get

$\rm \:  =  \: secA - \dfrac{1}{secA - tanA} \times \dfrac{secA + tanA}{secA + tanA}$

$\rm \:  =  \: secA - \dfrac{secA + tanA}{ {sec}^{2} A - {tan}^{2} A}$

$\rm \:  =  \: secA - (secA + tanA)$

$\rm \:  =  \: secA - secA - tanA$

$\rm \:  =  \: - \: tanA$

So,

$\rm :\longmapsto\:\boxed{ \bf{ \:\dfrac{1}{cosA} - \dfrac{1}{secA + tanA} = - \: tanA}} - - (2)$

From equation (1) and (2), we concluded that

$\boxed{ \bf{ \: \frac{1}{secA + tanA} - \frac{1}{cosA} = \frac{1}{cosA} - \frac{1}{secA + tanA}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1