Interesting Trigonometry!!

Algebra Level 5

If $\alpha$ , $\beta$ , and $\gamma$ are acute angles, and

$\begin{cases} \cos \theta =\dfrac { \sin \beta}{ \sin \alpha} \\ \cos \varphi =\dfrac {\sin\gamma}{\sin \alpha} \\ \cos (\theta -\varphi) = \sin \beta \sin \gamma \end{cases}$

find the value of $\tan ^{ 2 }{ \alpha } -\tan ^{ 2 }{ \beta } -\tan ^{ 2 }{ \gamma }$ .

The answer is 0.

1 solution

De Silva
Jan 9, 2015

cos θ = sin B / sin a cos p = sin r/ sin a

sin θ = ( sin^2 a - sin^ B )^0.5 / sin a sin p = ( sin^2 a - sin^ r )^0.5 / sin a

cos ( θ- p ) = sin B . sin r

cos θ . cos p + sin θ . sin p = sin B . sin r

[ sin B / sin a ] . [sin r/ sin a ] + [ ( sin^2 a - sin^2 B )^0.5 / sin a ] . [ ( sin^2 a - sin^2 r )^0.5 / sin a ] =sin B . sin r

sin B . sin r + [ ( sin^2 a - sin^2 B )^0.5 ] . [ ( sin^2 a - sin^2 r )^0.5 ] = sin B . sin r . ( sin^ 2 r)

-( cos^2 a) . sin B . sin r = [ ( sin^2 a - sin^2 B )^0.5 ] . [ ( sin^2 a - sin^2 r )^0.5 ]

sin B . sin r = { [ ( sin^2 a - sin^2 B ) / cos^2 a ]^0.5 } .{ [ ( sin^2 a - sin^2 r ) / cos^2 a ]^0.5 }

sin^2 B . sin^2 r = [ tan^2 a - ( sin^2 B / cos^2 a ) ] . [ tan^2 a - ( sin^2 r/ cos^2 a ) ]

sin^2 B . sin^2 r = [ tan^2 a - ( 1 + tan^2 a ) . sin^2 B ] . [ [ tan^2 a - ( 1 + tan^2 a ) . sin^2 r ]

sin^2 B . sin^2 r = ( tan^2 a . cos^2 B - sin^2 B ) . ( tan^2 a . cos^2 r - sin^2 r)

1 = { ( tan^2 a / tan^2 B ) - 1 } . { ( tan^2 a / tan^2 r ) - 1 }

tan^2 a . ( tan^2 r + tan^2 B ) = tan^4 a

As a is an acute angle , a is non equal to zero ...... so tan a is non equal to zero

tan^2 r + tan^2 B = tan^2 a

tan^2 a - tan^2 B - tan^2 r = 0

Interesting Trigonometry!!

The answer is 0.

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