Question

1 + SinA - cosA / cosA- 1 + sinA = 1 - cosA / sinA.
Chapter trigonometry ratio..

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Answer 1

Correct Question :

$\sf \dfrac{1 + sin A - cos A}{cos A + 1 + sin A} = \dfrac{1 - cos A}{sin A}$

Answer :

To Prove :

• $\sf \dfrac{1 + sin A - cos A}{cos A + 1 + sin A} = \dfrac{1 - cos A}{sin A}$

Proof :

LHS = $\sf \dfrac{1 + sin A - cos A}{cos A + 1 + sin A}$

⠀ ⠀ ⠀⠀ ⠀

By dividing Numerator and Denominator by SinA :

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1 + sin A - cos A}{sinA}}{\dfrac{cos A + 1 + sin A}{sinA}}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1}{sinA} + \dfrac{sin A}{sinA} - \dfrac{cos A}{sinA}}{\dfrac{cos A}{sinA} + \dfrac{1}{sinA} + \dfrac{sin A}{sinA}}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1}{sinA} + \cancel{\dfrac{sin A}{sinA}} - \dfrac{cos A}{sinA}}{\dfrac{cos A}{sinA} + \dfrac{1}{sinA} + \cancel{\dfrac{sin A}{sinA}}}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1}{sinA} + 1 - \dfrac{cos A}{sinA}}{\dfrac{cos A}{sinA} + \dfrac{1}{sinA} + 1}$

⠀ ⠀ ⠀⠀ ⠀

We know that :

• $\large \underline{\boxed{\bf{\dfrac{1}{sin \theta} = cosec \theta }}}$

• $\large \underline{\boxed{\bf{\dfrac{cos \theta}{sin \theta} = cot \theta }}}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\pink{cosecA} + 1 - \pink{cotA}}{\pink{cotA} + \pink{cosecA} + 1}$

⠀ ⠀ ⠀⠀ ⠀

Now, we know that :

• $\large \underline{\boxed{\bf{1 = cosec^{2} \theta - cot^{2} \theta }}}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{cosecA + \pink{(cosec^{2} A - cot ^{2}A)} - cotA}{cotA + cosecA + 1}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{cosecA - cotA + (cosec^{2} A - cot ^{2}A)}{cotA + cosecA + 1}$

⠀ ⠀ ⠀⠀ ⠀

By using identinty :

• a² - b² = (a + b)(a - b)

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{cosecA - cotA + \pink{(cosec A + cotA)(cosec A - cotA)}}{cotA + cosecA + 1}$

⠀ ⠀ ⠀⠀ ⠀

Now, by taking cosecA - cotA common from numerator :

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\pink{(cosecA - cotA)} ( 1 + (cosec A + cotA)}{cotA + cosecA + 1}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{(cosecA - cotA) ( 1 + cosec A + cotA)}{(1 + cosec A + cotA)}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{(cosecA - cotA) \cancel{( 1 + cosec A + cotA)}}{\cancel{(1 + cosec A + cotA)}}$

$\sf : \implies \underline{\underline{\bf LHS}} = cosecA - cotA$

⠀ ⠀ ⠀⠀ ⠀

Now, we know that :

• $\large \underline{\boxed{\bf{cosec \theta = \dfrac{1}{sin \theta} }}}$

• $\large \underline{\boxed{\bf{cot \theta = \dfrac{cos \theta}{sin \theta} }}}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{1}{sin A} - \dfrac{cos A}{sin A}$

$\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{1 - cos A}{sin A}$

⠀ ⠀ ⠀⠀ ⠀

RHS = $\sf \dfrac{1 - cos A}{sin A}$

⠀ ⠀

As, LHS = RHS,

Hence, proved.

Answer 2

Answer:

Correct Question :

\sf \dfrac{1 + sin A - cos A}{cos A + 1 + sin A} = \dfrac{1 - cos A}{sin A}

cosA+1+sinA

1+sinA−cosA

=

sinA

1−cosA

Answer :

To Prove :

\sf \dfrac{1 + sin A - cos A}{cos A + 1 + sin A} = \dfrac{1 - cos A}{sin A}

cosA+1+sinA

1+sinA−cosA

=

sinA

1−cosA

Proof :

LHS = \sf \dfrac{1 + sin A - cos A}{cos A + 1 + sin A}

cosA+1+sinA

1+sinA−cosA

⠀ ⠀ ⠀⠀ ⠀

By dividing Numerator and Denominator by SinA :

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1 + sin A - cos A}{sinA}}{\dfrac{cos A + 1 + sin A}{sinA}}:⟹

LHS

=

sinA

cosA+1+sinA

sinA

1+sinA−cosA

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1}{sinA} + \dfrac{sin A}{sinA} - \dfrac{cos A}{sinA}}{\dfrac{cos A}{sinA} + \dfrac{1}{sinA} + \dfrac{sin A}{sinA}}:⟹

LHS

=

sinA

cosA

+

sinA

1

+

sinA

sinA

sinA

1

+

sinA

sinA

−

sinA

cosA

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1}{sinA} + \cancel{\dfrac{sin A}{sinA}} - \dfrac{cos A}{sinA}}{\dfrac{cos A}{sinA} + \dfrac{1}{sinA} + \cancel{\dfrac{sin A}{sinA}}}:⟹

LHS

=

sinA

cosA

+

sinA

1

+

sinA

sinA

sinA

1

+

sinA

sinA

−

sinA

cosA

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\dfrac{1}{sinA} + 1 - \dfrac{cos A}{sinA}}{\dfrac{cos A}{sinA} + \dfrac{1}{sinA} + 1}:⟹

LHS

=

sinA

cosA

+

sinA

1

+1

sinA

1

+1−

sinA

cosA

⠀ ⠀ ⠀⠀ ⠀

We know that :

\large \underline{\boxed{\bf{\dfrac{1}{sin \theta} = cosec \theta }}}

sinθ

1

=cosecθ

\large \underline{\boxed{\bf{\dfrac{cos \theta}{sin \theta} = cot \theta }}}

sinθ

cosθ

=cotθ

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\pink{cosecA} + 1 - \pink{cotA}}{\pink{cotA} + \pink{cosecA} + 1}:⟹

LHS

=

cotA+cosecA+1

cosecA+1−cotA

⠀ ⠀ ⠀⠀ ⠀

Now, we know that :

\large \underline{\boxed{\bf{1 = cosec^{2} \theta - cot^{2} \theta }}}

1=cosec

2

θ−cot

2

θ

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{cosecA + \pink{(cosec^{2} A - cot ^{2}A)} - cotA}{cotA + cosecA + 1}:⟹

LHS

=

cotA+cosecA+1

cosecA+(cosec

2

A−cot

2

A)−cotA

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{cosecA - cotA + (cosec^{2} A - cot ^{2}A)}{cotA + cosecA + 1}:⟹

LHS

=

cotA+cosecA+1

cosecA−cotA+(cosec

2

A−cot

2

A)

⠀ ⠀ ⠀⠀ ⠀

By using identinty :

a² - b² = (a + b)(a - b)

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{cosecA - cotA + \pink{(cosec A + cotA)(cosec A - cotA)}}{cotA + cosecA + 1}:⟹

LHS

=

cotA+cosecA+1

cosecA−cotA+(cosecA+cotA)(cosecA−cotA)

⠀ ⠀ ⠀⠀ ⠀

Now, by taking cosecA - cotA common from numerator :

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{\pink{(cosecA - cotA)} ( 1 + (cosec A + cotA)}{cotA + cosecA + 1}:⟹

LHS

=

cotA+cosecA+1

(cosecA−cotA)(1+(cosecA+cotA)

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{(cosecA - cotA) ( 1 + cosec A + cotA)}{(1 + cosec A + cotA)}:⟹

LHS

=

(1+cosecA+cotA)

(cosecA−cotA)(1+cosecA+cotA)

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{(cosecA - cotA) \cancel{( 1 + cosec A + cotA)}}{\cancel{(1 + cosec A + cotA)}}:⟹

LHS

=

(1+cosecA+cotA)

(cosecA−cotA)

(1+cosecA+cotA)

\sf : \implies \underline{\underline{\bf LHS}} = cosecA - cotA:⟹

LHS

=cosecA−cotA

⠀ ⠀ ⠀⠀ ⠀

Now, we know that :

\large \underline{\boxed{\bf{cosec \theta = \dfrac{1}{sin \theta} }}}

cosecθ=

sinθ

1

\large \underline{\boxed{\bf{cot \theta = \dfrac{cos \theta}{sin \theta} }}}

cotθ=

sinθ

cosθ

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{1}{sin A} - \dfrac{cos A}{sin A}:⟹

LHS

=

sinA

1

−

sinA

cosA

\sf : \implies \underline{\underline{\bf LHS}} = \dfrac{1 - cos A}{sin A}:⟹

LHS

=

sinA

1−cosA

⠀ ⠀ ⠀⠀ ⠀

RHS = \sf \dfrac{1 - cos A}{sin A}

sinA

1−cosA

⠀ ⠀

As, LHS = RHS,

Hence, proved.

Step-by-step explanation:

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