\(\tan \theta =\frac{2\mathrm x(\mathrm x+1)}{2\mathrm x+1}\)

\(\because \sec\theta =\sqrt{1+\tan^2\theta}\)

\(=\sqrt{1+\left(\frac{2\mathrm x(\mathrm x+1)}{2\mathrm x+1}\right)^2}\)

\(=\sqrt{\frac{(2\mathrm x+1)^2+(2\mathrm x(\mathrm x+1))^2}{2\mathrm x+1}}\)

\(=\frac{\sqrt{4\mathrm x^2+4\mathrm x+1+4\mathrm x^2(\mathrm x^2+2\mathrm x+1)}}{2\mathrm x+1}\)

\(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}{2\mathrm x+1}\)

\(\therefore\) \(\cos\theta = \frac{1}{\sec\theta}\) \(=\frac{2\mathrm x+1}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

Now, \(\sin \theta =\sqrt{1-\cos^2\theta}\)

\(=\sqrt{1-\frac{(2\mathrm x+1)^2}{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1-(4\mathrm x^2+4\mathrm x+1)}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{\sqrt{4\mathrm x^4+8\mathrm x^3+4\mathrm x^2}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)

\(=\frac{2\mathrm x\sqrt{\mathrm x^2+2\mathrm x+1}}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)    

\(=\frac{2\mathrm x(\mathrm x+1)}{\sqrt{4\mathrm x^4+8\mathrm x^3+8\mathrm x^2+4\mathrm x+1}}\)  \(\Big(\because (\mathrm x^2+2\mathrm x+1)^{\frac{1}{2}}\)\(=((\mathrm x+1)^2)^{\frac{1}{2}}= \mathrm x+1\Big)\)