(a)OR(b)

If $\displaystyle \sin \theta+\cos \theta=\sqrt{3}$, then prove that \[\tan \theta+\cot \theta=1 \]

\[\begin{aligned} & (\sin \theta+\cos \theta)^{2}=(\sqrt{3})^{2} \\ & \Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta=3 \\ & \Rightarrow \sin \theta \cos \theta=1 \quad--(\mathrm{i}) \\ & \text { LHS }=\tan \theta+\cot \theta=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \sin \theta}=\frac{1}{\cos \theta \sin \theta} \\ & \quad=1 \quad[\text { using (i)] } \\ & \quad=\text { RHS } \end{aligned} \]

\[\begin{aligned} \mathrm{LHS} & =\left(\sin \mathrm{A}+\frac{1}{\cos \mathrm{~A}}\right)^{2}+\left(\cos \mathrm{A}+\frac{1}{\sin \mathrm{~A}}\right)^{2} \\ & =\sin ^{2} \mathrm{~A}+\frac{1}{\cos ^{2} \mathrm{~A}}+\frac{2 \sin \mathrm{~A}}{\cos \mathrm{~A}}+\cos ^{2} \mathrm{~A}+\frac{1}{\sin ^{2} \mathrm{~A}}+\frac{2 \cos \mathrm{~A}}{\sin \mathrm{~A}} \\ & =1+\left(\frac{1}{\sin ^{2} \mathrm{~A}}+\frac{1}{\cos ^{2} \mathrm{~A}}\right)+2\left(\frac{\sin \mathrm{~A}}{\cos \mathrm{~A}}+\frac{\cos \mathrm{A}}{\sin \mathrm{~A}}\right) \\ & =1+\frac{1}{\cos ^{2} \mathrm{~A} \sin ^{2} \mathrm{~A}}+\frac{2}{\cos \mathrm{~A} \sin \mathrm{~A}} \\ & =1+\sec ^{2} \mathrm{~A} \operatorname{cosec}^{2} \mathrm{~A}+2 \sec \mathrm{~A} \operatorname{cosec} \mathrm{~A} \\ & =(1+\sec \mathrm{A} \operatorname{cosec} \mathrm{~A})^{2}=\mathrm{RHS} \end{aligned} \]