Question

If $7sin\alpha =24cos\alpha ;0<\alpha \frac{\pi }{2}$,then value of 14 $tan\alpha -75cos\alpha -7sec\alpha$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$Since0<\alpha <\frac{\pi }{2},allthetrigonometricalvalueswillbepositive.Given7sin\alpha =24cos\alpha or\frac{sin\alpha }{cos\alpha }=\frac{24}{7}\therefore tan\alpha =\frac{24}{7}---\left(1\right)squaring\left(1\right)weget{tan}^2\alpha ={\left(\frac{24}{7}\right)}^{2}or{sec}^2\alpha -1={\left(\frac{24}{7}\right)}^{2}or{sec}^2\alpha =1+\frac{{24}^2}{7^2}orsec\alpha =\sqrt{1+\frac{{24}^2}{7^2}}=\frac{25}{7}---\left(2\right)from\left(2\right)sec\alpha =\frac{25}{7}or\frac{1}{cos\alpha }=\frac{25}{7}\therefore cos\alpha =\frac{7}{25}---\left(3\right)Nowsubstitutingthevalueoftan\alpha ,cos\alpha andsec\alpha from\left(1\right),\left(2\right)and\left(3\right)weget14tan\alpha -75cos\alpha -7sec\alpha =14\times \frac{24}{7}-75\times \frac{7}{25}-7\times \frac{7}{25}=2\left(Ans\right)$