Question

tanA/1-cotA+cotA/1-tanA=sinAcosA+1/sinAcosA​

Answer 1

$\huge\red{\boxed{\sf AnweR}}$

Given:-

$\sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}=\dfrac{sinAcosA+1}{sinAcosA}$

Proof:-

• Convert this into SinA and CosA

$\bullet\sf tanA= \dfrac{sinA}{cosA}\ \ ;\ \ \bullet \sf cotA=\dfrac{cosA}{sinA}$

taking LHS:-

$\begin{gathered}\dashrightarrow\sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}\\ \\ \\ \dashrightarrow\sf \dfrac{\bigg\{\dfrac{sinA}{cosA}\bigg\}}{\bigg\{1-\dfrac{cosA}{sinA}\bigg\}}\ + \dfrac{\bigg\{\dfrac{cosA}{sinA}\bigg\}}{\bigg\{1-\dfrac{sinA}{cosA}\bigg\}}\\ \\ \\ \dashrightarrow\sf \dfrac{\bigg\{\dfrac{sinA}{cosA}\bigg\}}{\bigg\{\dfrac{sinA-cosA}{sinA}\bigg\}} \ + \dfrac{\bigg\{\dfrac{cosA}{sinA}\bigg\}}{\bigg\{\dfrac{cosA-sinA}{cosA}\bigg\}} \\ \\ \\ \dashrightarrow\sf \bigg\{\dfrac{sinA}{cosA}\bigg\}\times \bigg\{\dfrac{sinA}{sinA-cosA}\bigg\}\ + \bigg\{\dfrac{cosA}{sinA}\bigg\}\times \bigg\{\dfrac{cosA}{cosA-sinA}\bigg\}\\ \\ \\ \dashrightarrow\sf \bigg\{\dfrac{sin^2A}{cosA\big(sinA - cosA\big)}\bigg\}\ + \bigg\{\dfrac{cos^2A}{sinA\big(cosA-sinA\big)}\bigg\}\end{gathered}$

$\begin{gathered}\dashrightarrow\sf \bigg\{\dfrac{sin^2A}{cosA\big(sinA - cosA\big)}\bigg\} \ -\bigg\{ \dfrac{cos^2A}{sinA\big(sinA-cos A\big)}\bigg\}\\ \\ \\ \dashrightarrow\sf \dfrac{sin^3A-cos^3A}{sinA\ cosA\big(sinA-cosA\big)} \end{gathered}$

Formula Used Here :-

$\begin{gathered}\underline{\star{\scriptsize{\sf\ \ \ a^3-b^3= (a-b)(a^2+b^2+ab)}}}\\ \\ \\ \dashrightarrow\sf \dfrac{\cancel{\big(sinA-cosA\big)}\big(sin^2A+cos^2A+sinA. cosA\big)}{sinA cosA\cancel{\big(sinA-cosA\big)}}\\ \\ \\ \underline{\star{\scriptsize {\sf\ \ sin^2A+cos^2A=1}}}\\ \\ \\ \dashrightarrow\sf \bigg\lgroup\dfrac{1+ sinA \ cosA}{sinA\ cosA}\bigg\rgroup\ \ \ \ \ Hence\ Proved !!\end{gathered}$

Answer 2

Answer:

\huge\red{\boxed{\sf AnweR}}

AnweR

Given:-

\sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}=\dfrac{sinAcosA+1}{sinAcosA}

1−cotA

tanA

+

1−tanA

cotA

=

sinAcosA

sinAcosA+1

Proof:-

Convert this into SinA and CosA

\bullet\sf tanA= \dfrac{sinA}{cosA}\ \ ;\ \ \bullet \sf cotA=\dfrac{cosA}{sinA}∙tanA=

cosA

sinA

; ∙cotA=

sinA

cosA

taking LHS:-

\begin{gathered}\begin{gathered}\dashrightarrow\sf \dfrac{tanA}{1-cotA}+\dfrac{cotA}{1-tanA}\\ \\ \\ \dashrightarrow\sf \dfrac{\bigg\{\dfrac{sinA}{cosA}\bigg\}}{\bigg\{1-\dfrac{cosA}{sinA}\bigg\}}\ + \dfrac{\bigg\{\dfrac{cosA}{sinA}\bigg\}}{\bigg\{1-\dfrac{sinA}{cosA}\bigg\}}\\ \\ \\ \dashrightarrow\sf \dfrac{\bigg\{\dfrac{sinA}{cosA}\bigg\}}{\bigg\{\dfrac{sinA-cosA}{sinA}\bigg\}} \ + \dfrac{\bigg\{\dfrac{cosA}{sinA}\bigg\}}{\bigg\{\dfrac{cosA-sinA}{cosA}\bigg\}} \\ \\ \\ \dashrightarrow\sf \bigg\{\dfrac{sinA}{cosA}\bigg\}\times \bigg\{\dfrac{sinA}{sinA-cosA}\bigg\}\ + \bigg\{\dfrac{cosA}{sinA}\bigg\}\times \bigg\{\dfrac{cosA}{cosA-sinA}\bigg\}\\ \\ \\ \dashrightarrow\sf \bigg\{\dfrac{sin^2A}{cosA\big(sinA - cosA\big)}\bigg\}\ + \bigg\{\dfrac{cos^2A}{sinA\big(cosA-sinA\big)}\bigg\}\end{gathered}\end{gathered}

⇢

1−cotA

tanA

+

1−tanA

cotA

⇢

{1−

sinA

cosA

}

{

cosA

sinA

}

+

{1−

cosA

sinA

}

{

sinA

cosA

}

⇢

{

sinA

sinA−cosA

}

{

cosA

sinA

}

+

{

cosA

cosA−sinA

}

{

sinA

cosA

}

⇢{

cosA

sinA

}×{

sinA−cosA

sinA

} +{

sinA

cosA

}×{

cosA−sinA

cosA

}

⇢{

cosA(sinA−cosA)

sin

2

A

} +{

sinA(cosA−sinA)

cos

2

A

}

\begin{gathered}\begin{gathered}\dashrightarrow\sf \bigg\{\dfrac{sin^2A}{cosA\big(sinA - cosA\big)}\bigg\} \ -\bigg\{ \dfrac{cos^2A}{sinA\big(sinA-cos A\big)}\bigg\}\\ \\ \\ \dashrightarrow\sf \dfrac{sin^3A-cos^3A}{sinA\ cosA\big(sinA-cosA\big)} \end{gathered}\end{gathered}

⇢{

cosA(sinA−cosA)

sin

2

A

} −{

sinA(sinA−cosA)

cos

2

A

}

⇢

sinA cosA(sinA−cosA)

sin

3

A−cos

3

A

Formula Used Here :-

\begin{gathered}\begin{gathered}\underline{\star{\scriptsize{\sf\ \ \ a^3-b^3= (a-b)(a^2+b^2+ab)}}}\\ \\ \\ \dashrightarrow\sf \dfrac{\cancel{\big(sinA-cosA\big)}\big(sin^2A+cos^2A+sinA. cosA\big)}{sinA cosA\cancel{\big(sinA-cosA\big)}}\\ \\ \\ \underline{\star{\scriptsize {\sf\ \ sin^2A+cos^2A=1}}}\\ \\ \\ \dashrightarrow\sf \bigg\lgroup\dfrac{1+ sinA \ cosA}{sinA\ cosA}\bigg\rgroup\ \ \ \ \ Hence\ Proved !!\end{gathered}\end{gathered}

⋆ a

3

−b

3

=(a−b)(a

2

+b

2

+ab)

⇢

sinAcosA

(sinA−cosA)

(sinA−cosA)

(sin

2

A+cos

2

A+sinA.cosA)

⋆ sin

2

A+cos

2

A=1

⇢

⎩

⎪

⎪

⎪

⎧

sinA cosA

1+sinA cosA

⎭

⎪

⎪

⎪

⎫

Hence Proved!!