Question

Prove that
(sin²A)/[cosA(sinA - cosA)] + (cos²A)/[sinA(cosA - sinA)]

= 1 + tanA + cotA
​

Answer 1

Prove that

$\sf\: \dfrac{ {sin}^{2}A }{cosA(sinA - cosA)} + \dfrac{ {cos}^{2} A}{sinA(cosA - sinA)} = 1 + tanA + cotA$

$\large\underline{\bold{Solution-}}$

Identities Used :-

$1. \: \: \boxed{ \bf{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )}}$

$2. \: \: \boxed{ \bf{tanA = \dfrac{sinA}{cosA} }}$

$3. \: \: \boxed{ \bf{cotA = \dfrac{cosA}{sinA} }}$

Consider, LHS

$\sf\: \dfrac{ {sin}^{2}A }{cosA(sinA - cosA)} + \dfrac{ {cos}^{2} A}{sinA(cosA - sinA)}$

$\sf\: = \dfrac{ {sin}^{2}A }{cosA(sinA - cosA)} - \dfrac{ {cos}^{2} A}{sinA(sinA - cosA)}$

$\sf\: =\dfrac{ {sin}^{3}A - {cos}^{3}A}{sinA \: cosA \: (sinA - cosA)}$

$\sf\: = \: \dfrac{ \cancel{(sinA - cosA)} \: \: ( {sin}^{2}A + sinAcosA + {cos}^{2}A}{sinA \: cosA \: \: \: \cancel{(sinA - cosA)}}$

$\sf\: = \: \dfrac{ {sin}^{2}A \: + \: sinA \: cosA \: + \: {cos}^{2}A}{sinA \: cosA}$

$\sf\: = \: \dfrac{ {sin}^{2} A}{sinAcosA} + \dfrac{sinAcosA}{sinAcosA} + \dfrac{ {cos}^{2}A }{sinAcosA}$

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

$\sf\: =tanA + 1 + cotA$

$\sf\: = \: 1 + tanA + cotA$

$\sf\: =RHS$

$\bf{Hence, \: Proved}$

Additional Information :-

$1. \: \: \boxed{ \bf{cosecA \: = \: \dfrac{1}{sinA} }}$

$2. \: \: \boxed{ \bf{secA \: = \: \dfrac{1}{cosA} }}$

$3. \: \: \boxed{ \bf{ {sin}^{2}A + {cos}^{2}A = 1}}$

$4. \: \: \boxed{ \bf{ {cosec}^{2} A - {cot}^{2} A = 1}}$

$5. \: \: \boxed{ \bf{ {sec}^{2}A - {tan}^{2} A = 1}}$

Answer 2

Answer:

Prove that

\sf\: \dfrac{ {sin}^{2}A }{cosA(sinA - cosA)} + \dfrac{ {cos}^{2} A}{sinA(cosA - sinA)} = 1 + tanA + cotA

cosA(sinA−cosA)

sin

2

A

+

sinA(cosA−sinA)

cos

2

A

=1+tanA+cotA

\large\underline{\bold{Solution-}}

Solution−

Identities Used :-

1. \: \: \boxed{ \bf{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )}}1.

x

3

−y

3

=(x−y)(x

2

+xy+y

2

)

2. \: \: \boxed{ \bf{tanA = \dfrac{sinA}{cosA} }}2.

tanA=

cosA

sinA

3. \: \: \boxed{ \bf{cotA = \dfrac{cosA}{sinA} }}3.

cotA=

sinA

cosA

Consider, LHS

\sf\: \dfrac{ {sin}^{2}A }{cosA(sinA - cosA)} + \dfrac{ {cos}^{2} A}{sinA(cosA - sinA)}

cosA(sinA−cosA)

sin

2

A

+

sinA(cosA−sinA)

cos

2

A

\sf\: = \dfrac{ {sin}^{2}A }{cosA(sinA - cosA)} - \dfrac{ {cos}^{2} A}{sinA(sinA - cosA)}=

cosA(sinA−cosA)

sin

2

A

−

sinA(sinA−cosA)

cos

2

A

\sf\: =\dfrac{ {sin}^{3}A - {cos}^{3}A}{sinA \: cosA \: (sinA - cosA)}=

sinAcosA(sinA−cosA)

sin

3

A−cos

3

A

\sf\: = \: \dfrac{ \cancel{(sinA - cosA)} \: \: ( {sin}^{2}A + sinAcosA + {cos}^{2}A}{sinA \: cosA \: \: \: \cancel{(sinA - cosA)}}=

sinAcosA

(sinA−cosA)

(sinA−cosA)

(sin

2

A+sinAcosA+cos

2

A

\sf\: = \: \dfrac{ {sin}^{2}A \: + \: sinA \: cosA \: + \: {cos}^{2}A}{sinA \: cosA}=

sinAcosA

sin

2

A+sinAcosA+cos

2

A

\sf\: = \: \dfrac{ {sin}^{2} A}{sinAcosA} + \dfrac{sinAcosA}{sinAcosA} + \dfrac{ {cos}^{2}A }{sinAcosA}=

sinAcosA

sin

2

A

+

sinAcosA

sinAcosA

+

sinAcosA

cos

2

A

\sf\: = \: \dfrac{sinA}{cosA} + 1 + \dfrac{cosA}{sinA}=

cosA

sinA

+1+

sinA

cosA

\sf\: =tanA + 1 + cotA=tanA+1+cotA

\sf\: = \: 1 + tanA + cotA=1+tanA+cotA

\sf\: =RHS=RHS

\bf{Hence, \: Proved}Hence,Proved

Additional Information :-

1. \: \: \boxed{ \bf{cosecA \: = \: \dfrac{1}{sinA} }}1.

cosecA=

sinA

1

2. \: \: \boxed{ \bf{secA \: = \: \dfrac{1}{cosA} }}2.

secA=

cosA

1

3. \: \: \boxed{ \bf{ {sin}^{2}A + {cos}^{2}A = 1}}3.

sin

2

A+cos

2

A=1

4. \: \: \boxed{ \bf{ {cosec}^{2} A - {cot}^{2} A = 1}}4.

cosec

2

A−cot

2

A=1

5. \: \: \boxed{ \bf{ {sec}^{2}A - {tan}^{2} A = 1}}5.

sec

2

A−tan

2

A=1