Question

If 0<$\theta$<$\frac{\mathrm{\pi }}{2}$and sin $\theta$ + cos ​$\theta$ +tan ​$\theta$+cot ​$\theta$ +sec ​$\theta$ +cosec ​$\theta$= 7, then show that sin 2​$\theta$ is a root of the equation $x^2$-44x+36 =0

Answer 1

Dear student

$$\begin{gathered} \mathrm{sin\theta }+\mathrm{cos\theta }+\mathrm{tan\theta }+\mathrm{cot\theta }+\mathrm{sec\theta }+\mathrm{cosec\theta }=7 \\ \Rightarrow \mathrm{sin\theta }+\mathrm{cos\theta }+\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}+\frac{1}{\mathrm{cos\theta }}+\frac{1}{\mathrm{sin\theta }}=7 \\ \Rightarrow \frac{{\sin }^2\mathrm{\theta }\mathrm{cos\theta }+{\cos }^2\mathrm{\theta sin\theta }+{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }+\mathrm{sin\theta }+\mathrm{cos\theta }}{\mathrm{sin\theta }\mathrm{cos\theta }}=7 \\ \Rightarrow \mathrm{sin\theta }\mathrm{cos\theta }(\mathrm{sin\theta }+\mathrm{cos\theta })+1+\mathrm{sin\theta }+\mathrm{cos\theta }=7\mathrm{sin\theta }\mathrm{cos\theta } \\ \mathrm{Let}\mathbf{sin\theta }\mathbf{}\mathbf{cos\theta }\mathbf{=}\mathbf{v} \\ \mathrm{and}\frac{1}{2}\times 2\mathrm{sin\theta }\mathrm{cos\theta }=\mathrm{v} \\ \Rightarrow \frac{1}{2}\sin 2\mathrm{\theta }=\mathrm{v} \\ \Rightarrow \mathbf{sin}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{2}\mathbf{v}...\left(\mathrm{A}\right) \\ \mathrm{and}\mathbf{sin\theta }\mathbf{+}\mathbf{cos\theta }\mathbf{=}\mathbf{u} \\ \Rightarrow {\mathrm{u}}^2=(\mathrm{sin\theta }+\mathrm{cos\theta }{)}^2={\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }+2\mathrm{sin\theta }\mathrm{cos\theta }=1+2\mathrm{sin\theta }\mathrm{cos\theta }=1+2\mathrm{v}...(1) \\ \mathrm{So},\mathrm{uv}+1+\mathrm{u}=7\mathrm{v} \\ \Rightarrow \mathrm{u}(\mathrm{v}+1)=7\mathrm{v}-1 \\ \Rightarrow \mathrm{u}=\frac{7\mathrm{v}-1}{\mathrm{v}+1} \\ \Rightarrow {\mathrm{u}}^2={\left(\frac{7\mathrm{v}-1}{\mathrm{v}+1}\right)}^2 \\ \Rightarrow (1+2\mathrm{v})={\left(\frac{7\mathrm{v}-1}{\mathrm{v}+1}\right)}^2..(2) \\ \mathrm{Let}\mathbf{sin}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{v}[\mathrm{using}\mathrm{A}] \\ \mathrm{Then},(2)\mathrm{becomes} \\ \Rightarrow 1+\mathrm{x}=\frac{{\left({\displaystyle \frac{7\mathrm{x}}{2}}-1\right)}^2}{{\left({\displaystyle \frac{\mathrm{x}}{2}}+1\right)}^2} \\ \Rightarrow 1+\mathrm{x}=\frac{{\displaystyle \frac{49{\mathrm{x}}^2}{4}}+1-7\mathrm{x}}{{\displaystyle \frac{{\mathrm{x}}^2}{4}}+1+\mathrm{x}} \\ \Rightarrow 1+\mathrm{x}=\frac{49{\mathrm{x}}^2+4-28\mathrm{x}}{{\mathrm{x}}^2+4+4\mathrm{x}} \\ \Rightarrow {\mathrm{x}}^2+4\mathrm{x}+4+{\mathrm{x}}^3+4\mathrm{x}+4{\mathrm{x}}^2=49{\mathrm{x}}^2+4-28\mathrm{x} \\ \Rightarrow {\mathrm{x}}^3-44{\mathrm{x}}^2+36\mathrm{x}=0 \\ \Rightarrow {\mathrm{x}}^2-44\mathrm{x}+36=0 \\ \mathrm{So},\sin 2\mathrm{\theta }\mathrm{is}\mathrm{a}\mathrm{root}\mathrm{of}{\mathrm{x}}^2-44\mathrm{x}+36=0 \end{gathered}$$
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