If  7 sin α =24 cos α were 0<α<π2, then value of 14 tan α -75cos α -7sec α is equals to

# If  were $0<\alpha <\frac{\pi }{2}$, then value of is equals to

1. A
$1$
2. B
$2$
3. C
$3$
4. D
$4$

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### Solution:

Given,
$7\mathit{sin}\alpha =24\mathit{cos}\alpha$
Divide both side of this equation by $7\mathit{cos}\alpha$,
$\frac{\mathit{sin}\alpha }{\mathit{cos}\alpha }=\frac{24}{7}$
$\mathit{tan}\alpha =\frac{24}{7}$     …. (1)           ()
It is known that,
${\mathit{sec}}^{2}\alpha -{\mathit{tan}}^{2}\alpha =1$
From equation 1,
${\mathit{sec}}^{2}\alpha -{\left(\frac{24}{7}\right)}^{2}=1$
${\mathit{sec}}^{2}\alpha =1+{\left(\frac{24}{7}\right)}^{2}$
${\mathit{sec}}^{2}\alpha =\frac{49+576}{49}$
${\mathit{sec}}^{2}\alpha =\frac{625}{49}$T
$\mathit{sec}\alpha =\sqrt{\frac{625}{49}}$
$\mathit{sec}\alpha =\frac{25}{7}$
$\mathit{cos}\alpha =\frac{7}{25}$                     Thus,
,
and
$\mathit{tan}\alpha =\frac{24}{7}$
Now put the values in the given equation,
$14\mathit{tan}\alpha -75\mathit{cos}\alpha -7\mathit{sec}\alpha$
$=14×\frac{24}{7}-75×\frac{7}{25}-7×\frac{25}{7}$
$=48-21-25$
$=2$
Hence, option 2 is correct.

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