JEE Main 2024 — Mathematics Trigonometry

Given: $4\mathrm{cos}\theta +5\mathrm{sin}\theta =1$4cosθ+5sinθ=1

$$\Rightarrow 4(\frac{1-{\mathrm{tan}}^{2}\frac{\theta }{2}}{1+{\mathrm{tan}}^{2}\frac{\theta }{2}})+5(\frac{2\mathrm{tan}\frac{\theta }{2}}{1+{\mathrm{tan}}^{2}\frac{\theta }{2}})=1$$⇒4(1+tan22θ​1−tan22θ​​)+5(1+tan22θ​2tan2θ​​)=1

$$\Rightarrow 5{\mathrm{tan}}^{2}\frac{\theta }{2}-10\mathrm{tan}\frac{\theta }{2}-3=0$$⇒5tan22θ​−10tan2θ​−3=0

$$\Rightarrow \mathrm{tan}\frac{\theta }{2}=\frac{10\pm \sqrt{100-4(5)(-3)}}{2\times 5}$$⇒tan2θ​=2×510±100−4(5)(−3)​​

$$\Rightarrow \mathrm{tan}\frac{\theta }{2}=\frac{10\pm \sqrt{160}}{10}$$⇒tan2θ​=1010±160​​

$$\Rightarrow \mathrm{tan}\frac{\theta }{2}=\frac{5\pm \sqrt{40}}{5}$$⇒tan2θ​=55±40​​

It is given that, $\alpha \in (\frac{-\pi }{2},\frac{\pi }{2})$α∈(2−π​,2π​)

$$\Rightarrow \frac{\alpha }{2}\in (\frac{-\pi }{4},\frac{\pi }{4})$$⇒2α​∈(4−π​,4π​)

$$\Rightarrow \mathrm{tan}\frac{\alpha }{2}\in (-1,1)$$⇒tan2α​∈(−1,1)

$$\Rightarrow \mathrm{tan}\frac{\alpha }{2}=\frac{5-\sqrt{40}}{5}$$⇒tan2α​=55−40​​

We know that, $\mathrm{tan}\theta =\frac{2\mathrm{tan}\frac{\theta }{2}}{1-{\mathrm{tan}}^{2}\frac{\theta }{2}}$tanθ=1−tan22θ​2tan2θ​​

$$\Rightarrow \mathrm{tan}\alpha =\frac{2(\frac{5-\sqrt{40}}{5})}{1-{(\frac{5-\sqrt{40}}{5})}^{2}}$$⇒tanα=1−(55−40​​)22(55−40​​)​

$$\Rightarrow \mathrm{tan}\alpha =\frac{2(\frac{5-\sqrt{40}}{5})}{2(\frac{\sqrt{40}}{5})-\frac{8}{5}}$$⇒tanα=2(540​​)−58​2(55−40​​)​

$$\Rightarrow \mathrm{tan}\alpha =\frac{5-\sqrt{40}}{\sqrt{40}-4}$$⇒tanα=40​−45−40​​

$$\Rightarrow \mathrm{tan}\alpha =\frac{\sqrt{40}-20}{24}$$⇒tanα=2440​−20​

$$\Rightarrow \mathrm{tan}\alpha =\frac{\sqrt{10}-10}{12}$$⇒tanα=1210​−10​