Question 56

If $$\sin(20 + x)^\circ = \cos 60^\circ, 0 \leq (20 + x) \leq 90$$, then find the value of $$2 \sin^2(3x + 15)^\circ - \cosec^2(2x + 10)^\circ$$.

Solution

$$\sin(20+x)^{\circ}=\cos60^{\circ}$$

$$\sin(20+x)^{\circ}=\sin30^{\circ}$$

$$20+x=30$$

$$x=10$$

$$2\sin^2(3x+15)^{\circ}-\operatorname{cosec}^2(2x+10)^{\circ}=2\sin^245^{\circ}-\operatorname{cosec}^230^{\circ}$$

$$=2\left(\frac{1}{\sqrt{2}}\right)^2-\left(2\right)^2$$

$$=2\left(\frac{1}{2}\right)-4$$

$$=1-4$$

$$=-3$$

Hence, the correct answer is Option B

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