Question

if tan beta/2= 4 tan alpha/2, prove that tan (beta-alpha)/2 = (3sin alpha)/ 5 - 3 cos aplha​

Answer 1

Question:

If $tan \frac{\beta} {2} = 4 tan \frac{ \alpha} {2}$, prove that, $tan \frac{\beta-\alpha}{2}= \frac{3 sin~\alpha} {5-3 cos \alpha}$

Solution:

We have a formula : $\Bigg\{ tan( \alpha-\beta) = \frac{tan \alpha - tan \beta}{1+tan \alpha ~ tan\beta} \Bigg\}$

Now, come to question,

We have:

$tan \frac{\beta-\alpha}{2} \\\\ tan \frac{ \beta} {2}- \frac{ \alpha} {2} \\\\ \sf Using~above~formula \\\\ tan \frac{ \beta} {2}- \frac{ \alpha} {2} = \frac{tan \frac{ \beta} {2} - tan \frac{ \alpha} {2}}{1+ tan \frac{ \beta} {2} ~tan\frac{ \alpha} {2} }- - - - [i]$

Putting $tan \frac{\beta} {2} = 4 tan \frac{ \alpha} {2}$, in [i], we have:

$\implies \frac{4 tan \frac{ \alpha} {2} - tan \frac{ \alpha} {2}}{1+4 tan \frac{ \alpha} {2} ~tan\frac{ \alpha} {2} } \\\\ \implies \frac{3 tan \frac{ \alpha} {2}} {1+ 4tan^2 \bigg(\frac{ \alpha} {2} \bigg)}$

Converting tan in terms of sin and cos:

$\frac{3 tan \frac{ \alpha} {2}} {1+ 4tan^2 \bigg(\frac{ \alpha} {2} \bigg)} \\ \\ \implies \frac{3 \frac{sin\frac{ \alpha} {2}}{cos\frac{ \alpha} {2}}} {1+4 \frac{sin^{2} \frac{ \alpha} {2}}{cos^{2} \frac{ \alpha}{2}}} \\\\ \implies \frac{3 \frac{sin\frac{ \alpha} {2}}{cos\frac{ \alpha} {2}}} { \frac{ cos^{2} \frac{ \alpha}{2}+ 4sin^{2} \frac{ \alpha} {2}}{cos^{2} \frac{ \alpha}{2}}}$

$\implies \: \frac{ \frac{3sin\frac{ \alpha} {2}}{cos\frac{ \alpha} {2}}} { \frac{ cos^{2} \frac{ \alpha}{2}+ 4sin^{2} \frac{ \alpha} {2}}{cos^{2} \frac{ \alpha}{2}}} \\ \\ \implies \frac{ \frac{3sin\frac{ \alpha} {2}}{cos\frac{ \alpha} {2}} \times \: cos^{2} \frac{ \alpha}{2}} { cos^{2} \frac{ \alpha}{2}+ 4sin^{2} \frac{ \alpha} {2}} \\ \\ \implies \frac{ 3sin \frac{ \alpha}{2} \times \: cos \frac{ \alpha}{2} }{ cos^{2} \frac{ \alpha}{2}+ 4sin^{2} \frac{ \alpha}{2}}$

Multiplying both numerator and denominator by 2.

$\implies \frac{ 2 \times 3sin \frac{ \alpha}{2} \times \: cos \frac{ \alpha}{2} }{ 2 cos^{2} \frac{ \alpha}{2}+ 8sin^{2} \frac{ \alpha}{2}}$

Here we will use three formulas which is as follows:

• $2sinAcosA= sin2A$
• $2cos^2 \frac{A}{2}= 1+cosA$
• $2sin^2 \frac{A} {2}= 1- cosA$

$\implies \frac{ 2 \times 3sin \frac{ \alpha}{2} \times \: cos \frac{ \alpha}{2} }{ 2 cos^{2} \frac{ \alpha}{2}+ 8sin^{2} \frac{ \alpha}{2}} \\\\ \implies \frac{3\times 2sin \frac{ \alpha}{2} \times \: cos \frac{ \alpha}{2}} { 2 cos^{2} \frac{ \alpha}{2}+ 4 \bigg(2sin^{2} \frac{ \alpha}{2} \bigg)} \\\\ \implies \frac{3sin \bigg( \frac{\alpha}{2} \times 2 \bigg)}{1+cos \alpha +4 (1- cos \alpha) } \\\\ \implies \frac{3sin \alpha}{1+cos \alpha +4 (1- cos \alpha) } \\\\ \implies \frac{3sin \alpha}{1+cos \alpha +4- 4cos \alpha } \\\\ \implies \frac{3 sin \alpha} {5-3cos \alpha}$