Question 45

If sec (3x - 20°) = cosec (3y + 20°), then what is the value of tan (x + y)?

Solution

Given : $$sec(3x-20^\circ)=cosec(3y+20^\circ)$$

=> $$sec(3x-20^\circ)=cosec[90^\circ-(-3y+70^\circ)]$$

$$\because cosec(90^\circ-\theta)=sec\theta$$

=> $$sec(3x-20^\circ)=sec(-3y+70^\circ)$$

=> $$3x-20^\circ=-3y+70^\circ$$

=> $$3x+3y=70+20=90^\circ$$

=> $$x+y=\frac{90}{3}=30^\circ$$ -----------(i)

$$\therefore$$ $$tan(x+y)=tan(30^\circ)=\frac{1}{\sqrt3}$$

=> Ans - (C)

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