Question

tanA/2 - cotA/2 +2cotA = 0​

Answer 1

Step-by-step explanation:

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tanA/2 - cotA/2+2cotA

$= \frac{ \sin( \frac{a}{2} ) }{ \cos( \frac{a}{2} ) } - \frac{ \cos( \frac{a}{2} ) }{ \sin( \frac{a}{2} ) } + 2 \frac{ \cos(a) }{ \sin(a) }$

$= \frac{ { \sin }^{2} \frac{a}{2} - { \cos }^{2} \frac{a}{2} }{ \sin( \frac{a}{2} ) \cos( \frac{a}{2} ) } + 2\frac{ { \cos }^{2} \frac{a}{2} - { \sin}^{2} \frac{a}{2} }{ 2\sin( \frac{a}{2} ) \cos( \frac{a}{2} ) }$

$= - (\frac{ { \cos }^{2} \frac{a}{2} - { \sin }^{2} \frac{a}{2} }{ \sin( \frac{a}{2} ) \cos( \frac{a}{2} ) } ) + \frac{ { \cos }^{2} \frac{a}{2} - { \sin}^{2} \frac{a}{2} }{ \sin( \frac{a}{2} ) \cos( \frac{a}{2} ) }$

= 0

Answer 2

Step-by-step explanation:

we have to prove that

tanA/2-cotA/2+2cotA=0

= LHS

tanA/2-cotA/2+2cotA

= {sin(A/2)/cos(A/2)} - {cos(A/2)/sin(A/2)} + 2cotA

={sin^2(A/2)-cos^2(A/2)} / {cos(A/2)*sin(A/2)}+2cotA

= -{cos^2(A/2)-sin^2(A/2)} / {cos(A/2)*sin(A/2)}+2cotA

= - {cosA} / {cos(A/2)*sin(A/2)+2cotA

since,cos^2A-sin^2A=cos2A

= - {cos(2A/2)} / {cos(A/2)*sin(A/2)} + 2cotA

= - {2*cosA} / {2*cos(A/2)*sin(A/2)} + 2cotA

= - {2cosA} /{sin(2A/2)}+2cotA

= {-(2cosA)/(sinA)}+2cotA

(since,sin2A=2*sinA*cosA)

= -2cotA+2cotA

= 0

= RHS