Question

Prove that cos inverse cos alpha + cos theta upon 1 + cos alpha cos theta is equal to 2 tan inverse tan alpha upon to 10 beta upon 2

Answer 1

Answer:

Step-by-step explanation:

RHS = 2 tan inverse tan alpha upon to 10 beta upon 2

=$2 tan^{-1} [ tan (\frac{\alpha}{2}) . tan \frac{\beta}{2}) ]$

= cos^{-1} $[ \frac{1-tan^{2} (\frac{\alpha}{2}) . tan^{2} (\frac{\beta}{2})}{1+tan^{2} (\frac{\alpha}{2}) . tan^{2} (\frac{\beta}{2})} ]$

= cos^{-1} [ $\frac{cos^{2}(\frac{\alpha}{2}).cos^{2}(\frac{\beta}{2})-sin^{2}(\frac{\alpha}{2}).sin^{2}(\frac{\beta}{2})}{cos^{2}(\frac{\alpha}{2}).cos^{2}(\frac{\beta}{2})+sin^{2}(\frac{\alpha}{2}).sin^{2}(\frac{\beta}{2})}$]

= cos^{-1} [$\frac{ { 2cos^{2}(\frac{\alpha}{2})}.{2cos^{2}(\frac{\beta}{2})} -{2sin^{2}(\frac{\alpha}{2})} - {2sin^{2}(\frac{\beta}{2})}}{ { 2cos^{2}(\frac{\alpha}{2})}.{2cos^{2}(\frac{\beta}{2})}+{2sin^{2}(\frac{\alpha}{2})} - {2sin^{2}(\frac{\beta}{2})}}$ ]

= cos^{-1} [$\frac{(1+cos \alpha ) (1+cos \beta)-( 1-cos \alpha ) (1+cos \beta)}{ (1+cos \alpha ) (1+cos \beta)+( 1-cos \alpha ) (1+cos \beta)}$ ]

= cos^{-1} [ $\frac{cos \alpha+cos\beta}{1+cos\alpha.cos\beta}$ ]

= LHS