Question

1. If cosec ø.cos(ø+ 54°) = 1, find the value of ø
so that ø and (ø + 54°) are acute angles.

2. if tan 4∅=cot 5∅, find ∅. given that both 4∅ and 5∅ are acute.
​

Answer 1

Hello mate,

Q1)

$\huge{\red{\star}} {\blue{\underline{\huge{\textsf{Solution}}}}} \huge{\red{\star}}$

$\csc( \alpha ) . \cos( 54 + \alpha ) = 1 \\ \\ \cos(54 + \alpha ) = \frac{1}{ \csc( \alpha ) } \\ \\ \cos( 54 + \alpha ) = \sin( \alpha ) \\ \\ \cos(90 - (36 - \alpha ) \: \: ) = \sin( \alpha ) \\ \\ we \: \: know \\ \\ \cos(90 - \alpha ) = \sin( \alpha ) \\ \\ so \\ \\ \cos(90 - (36 - \alpha ) \: \: ) \\ \\ \sin(36 - \alpha ) = \sin( \alpha ) \\ \\ 36 - \alpha = \alpha \\ \\ \alpha = 18$

Q2)

$\huge{\blue{\star}} {\red{\underline{\huge{\textsf{Solution}}}}} \huge{\blue{\star}}$

$\tan(4 \alpha ) = \cot( 5\alpha ) \\ \\ \tan(4 \alpha ) = \frac{1}{ \tan( 5\alpha ) } \\ \\ \tan(4 \alpha ) . \tan( 5\alpha ) = 1 \\ \\ \frac{ \sin( 4\alpha ) . \sin( 5\alpha ) }{ \cos( 4\alpha ). \cos( 5\alpha ) } = 1 \\ \\ \sin( 4\alpha ). \sin( 5\alpha ) = \cos( 4\alpha ). \cos( 5\alpha ) \\ \\ \sin( 4\alpha ). \sin( 5\alpha ) - \cos( 4\alpha ). \cos( 5\alpha ) = 0 \\ \\ \cos( 5\alpha + 3\alpha ) = 0 \\ \\ since \\ \\ \sin( \alpha ) . \sin( \beta ) - \cos( \alpha ) . \cos( \beta ) = \cos( \alpha + \beta ) \\ \\ \cos( 8\alpha ) = 0 \\ \\ we \: \: know \: \: \cos(90) = 0 \\ \\ \cos( 8\alpha ) = 0 \\ \\ 90 = 8\alpha \\ \\ \alpha = \frac{90}{8}$

Hope the solutions help you:)