Question

Prove that:
1 + secA - tanA / 1 + secA + tanA = 1- sinA / cosA​

Answer 1

Please refresh to the attachment for the answer of your question

Answer 2

TO PROVE :-

$\\ \sf \: \dfrac{1 + secA - tanA}{1 + secA + tanA} = \dfrac{1 - sinA}{cosA}$

IDENTITY USED :-

$\\ \sf \: 1) \: {sec}^{2}A - {tan}^{2}A = 1 \\ \\ \sf \: 2) \: secA = \dfrac{1}{cosA} \\ \\ \sf \: 3) \: tanA = \dfrac{sinA}{cosA} \\ \\ \sf \: 4) \: {x}^{2} - {y}^{2} = (x + y)(x - y)$

SOLUTION :-

$\\ \\ \sf \: L.H.S = \dfrac{1 + secA - tanA}{1 + secA + tanA}$

We know , sec²A - tan²A = 1

Substituting [ 1 = sec²A - tan²A ] in numerator ,

$\\ \\ \implies \sf \: \dfrac{( {sec}^{2}A - {tan}^{2}A ) + secA - tanA }{1 +secA + tanA}$

We know , x² - y² = (x+y)(x-y)

Here ,

• x = secA
• y = tanA

$\\ \\ \implies \sf \: \dfrac{(secA + tanA)(secA - tanA) + secA - tanA}{1 + secA + tanA}$

Now , we will take [secA-tanA] common in numerator.

$\\ \\ \implies \sf \: \dfrac{(secA - tanA) \cancel{ \{secA + tanA + 1 \}}}{ \cancel{1 + secA + tanA} } \\ \\ \\ \implies \sf \: secA - tanA$

We know ,

• secA = 1/cosA
• tanA = sinA/cosA

Substituting the values , we get..

$\\ \\ \implies \sf \: \dfrac{1}{cosA} - \dfrac{sinA}{cosA} \\ \\ \\ \implies \sf \: \dfrac{1 - sinA}{cosA} = R.H.S \: \: \: \: \: \: \: \: \: \: \: (proved)$

Hence ,

$\\ \underline{ \underline{\boxed{ \sf \: \dfrac{1 +sec -tan }{1 + sec +tan } = \dfrac{1 - sin}{cos} }}}$

MORE IDENTITIES :-

• sin²A + cos²A = 1

• 1 + cot²A = cosec²A

• sinA = 1/cosecA

• tanA = 1/cotA

• cotA = cotA/sinA