Question

TanA/secA-1-sinA/1+cosA=2cotA

Answer 1

Answer:

Explanation:

$\Large{\underline{\underline{\it{Given:}}}}$

$\sf{\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} =2\:cot\:A}$

$\Large{\underline{\underline{\it{To\:Prove:}}}}$

LHS = RHS

$\Large{\underline{\underline{\it{Solution:}}}}$

→ Taking the LHS of the equation,

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

→ Applying identities we get

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

→ Cross multiplying,

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

→ Cancelling cos A on both numerator and denominator

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

→ Again cross multiplying we get,

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

→ Taking sin A as common,

  $\sf{=\dfrac{sin\:A[1+cos\:A-(1-cos\:A)]}{(1^{2}-cos^{2}\:A ) }}$

  $\sf{=\dfrac{sin\:A[1+cos\:A-1+cos\:A]}{sin^{2}\:A } }$

→ Cancelling sin A on both numerator and denominator

  $\sf{=\dfrac{2\:cos\:A}{sin\:A} }$

 $\sf=2\times \dfrac{cos\:A}{sin\:A} }$

 $\sf{=2\:cot\:A}$

 $=\sf{RHS}$

→ Hence proved.

$\Large{\underline{\underline{\it{Identitites\:used:}}}}$

$\sf{tan\:A=\dfrac{sin\:A}{cos\:A} }$

$\sf{sec\:A=\dfrac{1}{cos\:A} }$

$\sf{(a+b)\times(a-b)=a^{2}-b^{2} }$

$\sf{(1-cos^{2}\:A)=sin^{2} \:A}$

$\sf{\dfrac{cos\:A}{sin\:A}=cot\:A}$

 

 

 

 

Answer 2

Answer:

Answer:

Explanation:

\Large{\underline{\underline{\it{Given:}}}}

Given:

\sf{\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} =2\:cot\:A}

secA−1

tanA

−

1+cosA

sinA

=2cotA

\Large{\underline{\underline{\it{To\:Prove:}}}}

ToProve:

LHS = RHS

\Large{\underline{\underline{\it{Solution:}}}}

Solution:

→ Taking the LHS of the equation,

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

secA−1

tanA

−

1+cosA

sinA

→ Applying identities we get

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

cosA

1

−1

cosA

sinA

−

1+cosA

sinA

→ Cross multiplying,

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

cosA

1−cosA

cosA

sinA

−

1+cosA

sinA

→ Cancelling cos A on both numerator and denominator

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

1−cosA

sinA

−

1+cosA

sinA

→ Again cross multiplying we get,

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

(1+cosA)(1−cosA)

sinA(1+cosA)−sinA(1−cosA)

→ Taking sin A as common,

\sf{=\dfrac{sin\:A[1+cos\:A-(1-cos\:A)]}{(1^{2}-cos^{2}\:A ) }}=

(1

2

−cos

2

A)

sinA[1+cosA−(1−cosA)]

\sf{=\dfrac{sin\:A[1+cos\:A-1+cos\:A]}{sin^{2}\:A } }=

sin

2

A

sinA[1+cosA−1+cosA]

→ Cancelling sin A on both numerator and denominator

\sf{=\dfrac{2\:cos\:A}{sin\:A} }=

sinA

2cosA

\sf=2\times \dfrac{cos\:A}{sin\:A} }

\sf{=2\:cot\:A}=2cotA

=\sf{RHS}=RHS

→ Hence proved.

\Large{\underline{\underline{\it{Identitites\:used:}}}}

Identititesused:

\sf{tan\:A=\dfrac{sin\:A}{cos\:A} }tanA=

cosA

sinA

\sf{sec\:A=\dfrac{1}{cos\:A} }secA=

cosA

1

\sf{(a+b)\times(a-b)=a^{2}-b^{2} }(a+b)×(a−b)=a

2

−b

2

\sf{(1-cos^{2}\:A)=sin^{2} \:A}(1−cos

2

A)=sin

2

A

\sf{\dfrac{cos\:A}{sin\:A}=cot\:A}

sinA

cosA

=cotA