If 7 sin α =24 cos α were 0

Given,

7 \sin α = 24 \cos α

Divide both side of this equation by

7 \cos α

\frac{\sin α}{\cos α} = \frac{24}{7}

\tan α = \frac{24}{7}

…. (1) (

since, \frac{\sin α}{\cos α} = \tan α

)
It is known that,

\sec^{2} α - \tan^{2} α = 1

From equation 1,

\sec^{2} α - {(\frac{24}{7})}^{2} = 1

\sec^{2} α = 1 + {(\frac{24}{7})}^{2}

\sec^{2} α = \frac{49 + 576}{49}

\sec^{2} α = \frac{625}{49}

\sec α = \sqrt{\frac{625}{49}}

\sec α = \frac{25}{7}

\cos α = \frac{7}{25}

(Since, \sec α = \frac{1}{\cos α})

Thus,

\sec α = \frac{25}{7}

\cos α = \frac{7}{25}

and

\tan α = \frac{24}{7}

Now put the values in the given equation,

14 \tan α - 75 \cos α - 7 \sec α

= 14 \times \frac{24}{7} - 75 \times \frac{7}{25} - 7 \times \frac{25}{7}

= 48 - 21 - 25

= 2

Hence, option 2 is correct.

If $7 \sin α = 24 \cos α$ were $0 < α < \frac{π}{2}$ , then value of $14 \tan α - 75 \cos α - 7 \sec α$ is equals to