JEE Main 2024 — Mathematics Trigonometry

Given: $\mathrm{cos}2x+a\cdot \mathrm{sin}x=2a-7$cos2x+a⋅sinx=2a−7

$$\Rightarrow 1-2{\mathrm{sin}}^{2}x+a\times \mathrm{sin}x=2a-7$$⇒1−2sin2x+a×sinx=2a−7

$$\Rightarrow a\times \mathrm{sin}x-2a=-8+2{\mathrm{sin}}^{2}x$$⇒a×sinx−2a=−8+2sin2x

$$\Rightarrow a(\mathrm{sin}x-2)=2({\mathrm{sin}}^{2}x-4)$$⇒a(sinx−2)=2(sin2x−4)

$$\Rightarrow a(\mathrm{sin}x-2)=2(\mathrm{sin}x-2)(\mathrm{sin}x+2)$$⇒a(sinx−2)=2(sinx−2)(sinx+2)

$$\Rightarrow \mathrm{sin}x=2[\text{Not possible}],a=2(\mathrm{sin}x+2)$$⇒sinx=2[Not possible],a=2(sinx+2)

We know that, $-1\leq \mathrm{sin}x\leq 1$−1≤sinx≤1

$$\Rightarrow a\in [2,6]$$⇒a∈[2,6]

$$\Rightarrow p=2,q=6$$⇒p=2,q=6

Now, $r=\mathrm{tan}{9}^{^{\circ}}+\mathrm{cot}{9}^{^{\circ}}-\mathrm{tan}{27}^{^{\circ}}-\mathrm{cot}{27}^{^{\circ}}$r=tan9∘+cot9∘−tan27∘−cot27∘

$$\Rightarrow r=\frac{{\mathrm{cos}}^{2}{9}^{^{\circ}}+{\mathrm{sin}}^{2}{9}^{^{\circ}}}{\mathrm{sin}{9}^{^{\circ}}\cdot \mathrm{cos}{9}^{^{\circ}}}-\frac{{\mathrm{cos}}^{2}{27}^{^{\circ}}+{\mathrm{sin}}^{2}{27}^{^{\circ}}}{\mathrm{sin}{27}^{^{\circ}}\cdot \mathrm{cos}{27}^{^{\circ}}}$$⇒r=sin9∘⋅cos9∘cos29∘+sin29∘​−sin27∘⋅cos27∘cos227∘+sin227∘​

$$\Rightarrow r=\frac{1}{\mathrm{sin}{9}^{^{\circ}}\cdot \mathrm{cos}{9}^{^{\circ}}}-\frac{1}{\mathrm{sin}{27}^{^{\circ}}\cdot \mathrm{cos}{27}^{^{\circ}}}$$⇒r=sin9∘⋅cos9∘1​−sin27∘⋅cos27∘1​

$$\Rightarrow r=\frac{2}{2\times \mathrm{sin}{9}^{^{\circ}}\cdot \mathrm{cos}{9}^{^{\circ}}}-\frac{2}{2\times \mathrm{sin}{27}^{^{\circ}}\cdot \mathrm{cos}{27}^{^{\circ}}}$$⇒r=2×sin9∘⋅cos9∘2​−2×sin27∘⋅cos27∘2​

$$\Rightarrow r=\frac{2}{\mathrm{sin}{18}^{^{\circ}}}-\frac{2}{\mathrm{sin}{54}^{^{\circ}}}$$⇒r=sin18∘2​−sin54∘2​

$$\Rightarrow r=2[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}]$$⇒r=2[5​−14​−5​+14​]

$$\Rightarrow r=8[\frac{\sqrt{5}+1-\sqrt{5}+1}{5-1}]$$⇒r=8[5−15​+1−5​+1​]

$$\Rightarrow r=8(\frac{2}{4})$$⇒r=8(42​)

$$\Rightarrow r=4$$⇒r=4

$$\Rightarrow p\cdot q\cdot r=2\times 6\times 4$$⇒p⋅q⋅r=2×6×4

$$\Rightarrow p\cdot q\cdot r=48$$⇒p⋅q⋅r=48