Question

Simplify:
1 + tan’A
1 + cotA​

Answer 1

Answer:

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Prove:

(1−cotA)

tanA

+

(1−tanA)

cotA

=secA⋅cosecA+1

Proof:

LHS

⇒

(1−cotA)

tanA

+

(1−tanA)

cotA

Putiing cotA=

tanA

1

⇒

(1−

tanA

1

)

tanA

+

(1−tanA)

tanA

1

⇒

(tanA−1)

tan

2

A

+

(1−tanA)

tanA

1

⇒

tanA−1

tan

2

A−

tanA

1

⇒

tanA(tanA−1)

tan

3

A−1

⇒

tanA(tanA−1)

(tanA−1)

3

+3tanA(tanA−1)

From (a−b)

3

=a

3

−b

3

−3ab(a−b)→a

3

−b

3

=(a−b)

3

+3ab(a−b)

Now,

tanA

(tanA−1)

2

+3tanA

tanA

tan

2

A+1−2tanA+3tanA

tanA

tan

2

A+1

+

tanA

tanA

sinAsecA

sec

2

A

+1

secA⋅cosecA+1=RHS