Question

Please prove the following compound identity
1/tanA -1/tan2A =cosec 2A

Answer 1

$\dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A$, proved.

Step-by-step explanation:

To prove the compound identity:

$\dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A$

L.H.S. = $\dfrac{1}{\tan A } -\dfrac{1}{\tan 2A }$

Using the trigonometric identity,

$\tan 2A = \dfrac{2\tan A}{1-\tan^2 A}$

= $\dfrac{1}{\tan A } -\dfrac{1}{\dfrac{2\tan A}{1-\tan^2 A}}$

$=\dfrac{1}{\tan A } -\dfrac{1-\tan^2 A}{2\tan A}$

Taking LCM of denominator part, we get

$=\dfrac{2-1+\tan^2 A}{2\tan A}$

$=\dfrac{1+\tan^2 A}{2\tan A}$

$=\dfrac{1}{\dfrac{2\tan A}{1+\tan^2 A} }$

Using the trigonometric identity,

$\sin 2A=\dfrac{2\tan A}{1+\tan^2 A}$

$=\dfrac{1}{\sin 2A}$

= $\csc 2A$

= R.H.S., proved.

∴ $\dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A$, proved.

Answer 2

Step-by-step explanation:

tanA

1

−

tan2A

1

=csc2A , proved.

Step-by-step explanation:

To prove the compound identity:

\dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A

tanA

1

−

tan2A

1

=csc2A

L.H.S. = \dfrac{1}{\tan A } -\dfrac{1}{\tan 2A }

tanA

1

−

tan2A

1

Using the trigonometric identity,

\tan 2A = \dfrac{2\tan A}{1-\tan^2 A}tan2A=

1−tan

2

A

2tanA

= \dfrac{1}{\tan A } -\dfrac{1}{\dfrac{2\tan A}{1-\tan^2 A}}

tanA

1

−

1−tan

2

A

2tanA

1

=\dfrac{1}{\tan A } -\dfrac{1-\tan^2 A}{2\tan A}=

tanA

1

−

2tanA

1−tan

2

A

Taking LCM of denominator part, we get

=\dfrac{2-1+\tan^2 A}{2\tan A}=

2tanA

2−1+tan

2

A

=\dfrac{1+\tan^2 A}{2\tan A}=

2tanA

1+tan

2

A

=\dfrac{1}{\dfrac{2\tan A}{1+\tan^2 A} }=

1+tan

2

A

2tanA

1

Using the trigonometric identity,

\sin 2A=\dfrac{2\tan A}{1+\tan^2 A}sin2A=

1+tan

2

A

2tanA

=\dfrac{1}{\sin 2A}=

sin2A

1

= \csc 2Acsc2A

= R.H.S., proved.

∴ \dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A

tanA

1

−

tan2A

1

=csc2A , proved