Given, tan θ + sec θ = 4

⇒ \(\frac{sin\,\theta}{cos\,\theta}\)+ \(\frac{1}{cos\,\theta} = 4\) \(\frac{1+sin\,\theta}{cos\,\theta}=4\)

⇒ \(\frac{sin^2\frac{\theta}{2}+cos^2\frac{\theta}{2}+2sin\frac{\theta}{2}cos\frac{\theta}{2}}{(cos^2\frac{\theta}{2}-sin^2\frac{\theta}{2})}=4\)

(Using the formulas sin² θ + cos² θ = 1, sin 2θ = 2sin θ cos θ, cos 2θ = cos² – sin²θ)

⇒ \(\frac{\bigg(sin\frac{\theta}{2}+cos\frac{\theta}{2}\bigg)^2}{\bigg(cos\frac{\theta}{2}+sin\frac{\theta}{2}\bigg)\bigg(cos\frac{\theta}{2}-sin\frac{\theta}{2}\bigg)}=4\)

⇒\(\frac{\bigg(sin\frac{\theta}{2}+cos\frac{\theta}{2}\bigg)}{cos\frac{\theta}{2}-sin\frac{\theta}{2}}=4\) ⇒ \(\frac{1+tan\,\frac{\theta}{2}}{1-tan\,\frac{\theta}{2}}=4\)

⇒ 1 + tan\(\frac{\theta}{2}\) = 4 – 4 tan\(\frac{\theta}{2}\) ⇒ 5 tan\(\frac{\theta}{2}\) = 3 ⇒ tan\(\frac{\theta}{2}\) = \(\frac35\)

∴ sin θ = \(\frac{2\,tan\,\frac{\theta}{2}}{1+tan^2\,\frac{\theta}{2}}\) = \(\frac{2\times\frac35}{1+\frac{9}{25}}\) = \(\frac{\frac65}{\frac{34}{25}}\) = \(\frac{30}{34}\) = \(\frac{15}{17}.\)

Alternatively, 

Given, tan θ + sec θ = 4                ...(i) 

sec²  = 1 + tan² θ 

⇒ sec² θ – tan² q = 1                 ...(ii) 

eq. (ii) ÷ eq. (i) 

⇒ sec θ – tan θ = \(\frac14\)         ...(iii) 

eq. (i) + eq. (iii) 

⇒ 2 sec θ = \(\frac{17}{4}\) ⇒ sec θ = \(\frac{17}{8}\) 

⇒ cos θ = \(\frac{17}{8}\)

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