Question

sin alpha sin beta sin Gama are in AP and cos alpha cos beta, gamma are in GP then cos square alpha + cos square gama - 4 cos alpha cos Gama by 1 - sin alpha sin Gama is equals to​

Answer 1

Step-by-step explanation:

cosα+cosβ+cosγ=0=sinα+sinβ+sinγ, prove that ... (iv) (sinθ+icosθ)n=[cos(π2−θ)+isin(π2− θ)]n=cosn(π2−θ)+isinn(π2−θ); eiθ=cosθ+isin ...

hope this helps ✌️✌️✌️

Answer 2

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Expand: $\frac{cos(α+γ)}{cos( \alpha - y)} </p><p></p><p>$

$\large{ \frac{cos( \alpha )cos(γ) - \sin( \alpha ) \sin( \gamma ) }{ \cos( \alpha ) \cos( \gamma ) + \sin( \alpha ) \sin( \gamma ) } }$

⚘ Now by Componendo-Dividendo

$\large{ \frac{ \cos( \alpha ) \cos( \gamma ) }{ \sin( \alpha ) \sin( \gamma ) } = \frac{1 + \cos(2 \beta ) }{1 - \cos( 2\beta ) } }$

⚘ We know that [Use C.D. after expanding ( ${cos\:2\beta}$ as ${cos\:(\beta)^{2}}$ −${sin\:(\beta)^{2}}$ )]

$\large{ \frac{1 + \cos( 2\beta ) }{1 - \cos( 2\beta ) } = \frac{1}{ \tan { (\beta })^{2} } }$

⚘ Dividing the numerator and denominator on the LHS by ${cos\:\alpha}$${sin\:\gamma}$

We get,

$\large{ \frac{ \cot( \gamma ) }{ \tan( \alpha ) } = \frac{1}{ \tan {( \beta })^{2} } }$