JEE Main 2021 — Mathematics Algebra

Let

$$z=\frac{1}{1-\mathrm{cos}\theta +2i\mathrm{sin}\theta }$$z=1−cosθ+2isinθ1​

$$\Rightarrow z=\frac{(1-\mathrm{cos}\theta )-2i\mathrm{sin}\theta }{(1-\mathrm{cos}\theta +2i\mathrm{sin}\theta )(1-\mathrm{cos}\theta -2i\mathrm{sin}\theta )}$$⇒z=(1−cosθ+2isinθ)(1−cosθ−2isinθ)(1−cosθ)−2isinθ​

$$\Rightarrow z=\frac{2{\mathrm{sin}}^{2}(\frac{\theta }{2})-2i\mathrm{sin}\theta }{(1-\mathrm{cos}\theta {)}^{2}+4{\mathrm{sin}}^{2}\theta }$$⇒z=(1−cosθ)2+4sin2θ2sin2(2θ​)−2isinθ​

Put,$\mathrm{sin}\theta =2\mathrm{sin}\frac{\theta }{2}\mathrm{cos}\frac{\theta }{2},1-\mathrm{cos}\theta =2{\mathrm{sin}}^{2}\frac{\theta }{2}$sinθ=2sin2θ​cos2θ​,1−cosθ=2sin22θ​

$$\Rightarrow z=\frac{2{\mathrm{sin}}^{2}(\frac{\theta }{2})-4i\mathrm{sin}(\frac{\theta }{2})\mathrm{cos}(\frac{\theta }{2})}{4{\mathrm{sin}}^{4}(\frac{\theta }{2})+16{\mathrm{sin}}^{2}(\frac{\theta }{2}){\mathrm{cos}}^{2}(\frac{\theta }{2})}$$⇒z=4sin4(2θ​)+16sin2(2θ​)cos2(2θ​)2sin2(2θ​)−4isin(2θ​)cos(2θ​)​

Hence,

$$Re(z)=\frac{2{\mathrm{sin}}^{2}(\frac{\theta }{2})}{4{\mathrm{sin}}^{4}(\frac{\theta }{2})+16{\mathrm{sin}}^{2}(\frac{\theta }{2}){\mathrm{cos}}^{2}(\frac{\theta }{2})}$$Re(z)=4sin4(2θ​)+16sin2(2θ​)cos2(2θ​)2sin2(2θ​)​

$$\Rightarrow \frac{1}{2{\mathrm{sin}}^{2}(\frac{\theta }{2})+8{\mathrm{cos}}^{2}(\frac{\theta }{2})}=\frac{1}{5}$$⇒2sin2(2θ​)+8cos2(2θ​)1​=51​

$$\Rightarrow \frac{1}{2{\mathrm{sin}}^{2}(\frac{\theta }{2})+8{\mathrm{cos}}^{2}(\frac{\theta }{2})}=\frac{1}{5}$$⇒2sin2(2θ​)+8cos2(2θ​)1​=51​

$$\Rightarrow {\mathrm{sin}}^{2}(\frac{\theta }{2})+4{\mathrm{cos}}^{2}(\frac{\theta }{2})=\frac{5}{2}$$⇒sin2(2θ​)+4cos2(2θ​)=25​

$$\Rightarrow 1-{\mathrm{cos}}^{2}(\frac{\theta }{2})+4{\mathrm{cos}}^{2}(\frac{\theta }{2})=\frac{5}{2}$$⇒1−cos2(2θ​)+4cos2(2θ​)=25​

$$\Rightarrow 3{\mathrm{cos}}^{2}\frac{\theta }{2}=\frac{3}{2}$$⇒3cos22θ​=23​

$$\Rightarrow {\mathrm{cos}}^{2}(\frac{\theta }{2})=\frac{1}{2}$$⇒cos2(2θ​)=21​

$$\Rightarrow \frac{\theta }{2}=n\pi \pm \frac{\pi }{4},n\in Z$$⇒2θ​=nπ±4π​,n∈Z

$$\Rightarrow \theta =2n\pi \pm \frac{\pi }{2}$$⇒θ=2nπ±2π​

For $\theta \in (0,\pi )$θ∈(0,π), $\theta =\frac{\pi }{2}$θ=2π​.

Hence,

$$\therefore {\int }_{0}^{\theta }\mathrm{sin}\theta d\theta ={\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}\theta d\theta ={[-\mathrm{cos}\theta ]}_{0}^{\frac{\pi }{2}}$$∴∫0θ​sinθdθ=∫02π​​sinθdθ=[−cosθ]02π​​

$$=-(0-1)$$=−(0−1)

$$=1$$=1