Question

1-sin A/cos A = cosA/1+sinA

Answer 1

GIVEN:

• 1-sin A/cos A = cosA/1+sinA

TO FIND:

• Evaluate

PROOF:

Taking L.H.S.

$\rm{\dashrightarrow \dfrac{(1-sinA)}{cosA} }$

Rationalising the denominator

$\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{cosA \times cos A} }$

$\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{cos^2 A} }$

Using Identity:-

• sin²A + cos²A = 1
• cos²A = 1 – sin²A

$\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{1-sin^2 A} }$

Using Identity:

• a² – b² = (a + b)(a – b)

$\rm{\dashrightarrow \dfrac{(1-sinA)cos A}{(1+sin A)(1-sin A)} }$

$\rm{\dashrightarrow \dfrac{\cancel{(1-sinA)}cos A}{(1+sin A)\cancel{(1-sin A)}} }$

$\bf{\dashrightarrow \dfrac{cos A}{1 + sin A} = R.H.S. }$

$\bf{\underline{\underline{ Hence \: proved}}}$

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Answer 2

Question:

Prove that,

${\rm{\dfrac{1 - \sin(A) }{\cos(A)} = \dfrac{\cos(A)}{1 + \sin(A)}}}$

To Find:

To prove that ,

${ \rm{ \dfrac{1 - \sin(A) }{\cos(A)} = \dfrac{\cos(A)}{1 + \sin(A)}}}$

Proof:

LHS = ${\rm{\dfrac{1 - \sin(A) }{\cos(A)}}}$

Multiplying numerator and denominator by cos(A)

We get,

$: \implies \rm \dfrac{[ 1 - \sin(A) ] \times \cos(A) }{\cos(A) \times \cos(A) }$

$: \implies \rm \dfrac{[ 1 - \sin(A) ] \times \cos(A) }{ { \cos}^{2} (A) }$

We know,

${\large{\green{\boxed{\pink{\star{\rm{{ \sin}^{2} (A) + { \cos}^{2} (A) = 1 }}}}}}}$

${\therefore{\pink{\underline{\green{\rm{{ \cos}^{2} (A) = 1 - { \sin}^{2} (A)}}}}}}$

$: \implies \rm \dfrac{[ 1 - \sin(A) ] \times \cos(A) }{ 1 - { \sin}^{2} (A) }$

We also know,

${\large{\green{\boxed{\pink{\star{\rm{{ ( a - b}^{2} = ( a + b )( a - b) }}}}}}}$

${\therefore{\pink{\underline{\green{\rm{1 - { \sin}^{2} (A) = [ 1 + \sin(A)][1 - \sin(A)}}}}}}$

$: \implies \rm \dfrac{[ 1 - \sin(A) ] \times \cos(A) }{ [ 1 + \sin(A)][1 - \sin(A)] }$

$: \implies \rm \dfrac{ \cancel{[ 1 - \sin(A) ] } \times \cos(A) }{ \cancel{[ 1 + \sin(A)]} [1 - \sin(A)] }$

$\therefore \rm \dfrac{ \cos(A) }{ [ 1 + \sin(A)] }$

HENCE IT IS PROVED .

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${\large{\purple{\underline{\underline{\red{\bf{Formulae\:used :}}}}}}}$

${\large{\pink{1)}}}$${\large{\green{\boxed{\orange{\star{\rm{{ \sin}^{2} (A) + { \cos}^{2} (A) = 1 }}}}}}}$

${\large{\pink{2)}}}$${\large{\green{\boxed{\orange{\star{\rm{{ ( a - b}^{2} = ( a + b )( a - b) }}}}}}}$