SSC CGL 6th March 2020 Shift-3 Question 75

Question 75

If A lies in third quadrant, and $$20 \tan A = 21$$, then the value of $$\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}$$

Solution

$$20 \tan A = 21$$

$$\tan A = 21/20$$

$$\frac{5 \sin A - 2 \cos A}{4 \cos A - \frac{5}{7} \sin A}$$

= $$\frac{\cos A(5 \frac{\sin A}{\cos A} - 2)}{\cos A(4 - \frac{5\sin A}{7\cos A})}$$

= $$\frac{5\tan A - 2}{4 - \frac{5}{7} \tan A}$$

On put the value of tan A,

= $$\frac{5 \times  \frac{21}{20}- 2}{4 - \frac{5}{7} \times \frac{21}{20}}$$

= $$\frac{\frac{13}{20}}{ \frac{13}{4}}$$ = $$\frac{1}{5}



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