Question

If
$\bf tan \; A = \dfrac{1- cos B}{sin \; B}$
then show that
$\bf{2A=tan B}$

Answer 1

Question:

$If \: \: \bf tan \; A = \dfrac{1- cos B}{sin \; B}$

$then, \: \bf{2A=tan B}$

$olution:

$\bf \: tan \: A = \frac{ \: 1 - cos \: B \: }{sin \: B \: }$

$\bf \: = > tan \: A \: = \frac{ \: 2 \: {sin }^{2 \: }( \frac{B}{2} )}{ \: 2 \: sin \: ( \frac{B}{2} ) \: cos \: ( \frac{B}{2} ) \: }$

$\bf = > tan \: A \: = \: tan \: (\frac{B}{2} )$

$\bf \: tan \: 2A = \frac{ \: 2 \: tan \: A \: }{ \: 1 - {tan}^{2} A \: } = \frac{ \: 2 \: tan \: (\frac{B}{2}) \: }{ \: 1 - \: {tan}^{2} (\frac{B}{2} ) \: }$

$\bf {.}^{ .}. \: \: tan \: 2A = tan \: B$

Answer 2

Step-by-step explanation:

Question:

If \: \: \bf tan \; A = \dfrac{1- cos B}{sin \; B}IftanA=

sinB

1−cosB

then, \: \bf{2A=tan B}then,2A=tanB

$olution:

\bf \: tan \: A = \frac{ \: 1 - cos \: B \: }{sin \: B \: }tanA=

sinB

1−cosB

\bf \: = > tan \: A \: = \frac{ \: 2 \: {sin }^{2 \: }( \frac{B}{2} )}{ \: 2 \: sin \: ( \frac{B}{2} ) \: cos \: ( \frac{B}{2} ) \: }=>tanA=

2sin(

2

B

)cos(

2

B

)

2sin

2

(

2

B

)

\bf = > tan \: A \: = \: tan \: (\frac{B}{2} )=>tanA=tan(

2

B

)

\bf \: tan \: 2A = \frac{ \: 2 \: tan \: A \: }{ \: 1 - {tan}^{2} A \: } = \frac{ \: 2 \: tan \: (\frac{B}{2}) \: }{ \: 1 - \: {tan}^{2} (\frac{B}{2} ) \: }tan2A=

1−tan

2

A

2tanA

=

1−tan

2

(

2

B

)

2tan(

2

B

)

\bf {.}^{ .}. \: \: tan \: 2A = tan \: B.

.

.tan2A=tanB