Question 55

If $$\sin^2 \theta = 2 \sin \theta - 1, 0^\circ \leq \theta \leq 90^\circ $$, then find the value of: $$\frac{1 + \cosec \theta}{1 - \cos \theta}$$.

Solution

$$\sin^2\theta=2\sin\theta-1$$

$$\sin^2\theta-2\sin\theta+1=0$$

$$\left(\sin\theta-1\right)^2=0$$

$$\sin\theta-1=0$$

$$\sin\theta=1$$

$$0^{\circ}\le\theta\le90^{\circ}$$

$$\Rightarrow$$ $$\theta=90^{\circ}$$

$$\frac{1+\operatorname{cosec}\theta}{1-\cos\theta}=\frac{1+\operatorname{cosec}90^{\circ\ }}{1-\cos90^{\circ\ }}$$

$$=\frac{1+1\ }{1-0\ }$$

$$=2$$

Hence, the correct answer is Option C

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