If tan$$\theta\ $$+ sec$$\ \theta\ $$= 3, $$\theta\ $$being acute, the value of 5 sin$$\theta\ $$ is:

Given : $$tan\theta+sec\theta=3$$

=> $$\frac{sin\theta}{cos\theta}+\frac{1}{cos\theta}=3$$

=> $$sin\theta+1=3cos\theta$$

Squaring both sides, we get :

=> $$sin^2\theta+1+2sin\theta=9cos^2\theta$$

=> $$sin^2\theta+1+2sin\theta=9(1-sin^2\theta)$$

=> $$sin^2\theta+1+2sin\theta=9-9sin^2\theta$$

=> $$10sin^2\theta+2sin\theta-8=0$$

Let $$sin\theta=x$$

=> $$5x^2+x-4=0$$

=> $$5x^2+5x-4x-4=0$$

=> $$5x(x+1)-4(x+1)=0$$

=> $$(x+1)(5x-4)=0$$

=> $$x=-1,\frac{4}{5}$$

$$\because \theta$$ is acute, => $$sin\theta\neq-1$$

$$\therefore$$ $$sin\theta=\frac{4}{5}$$

=> Ans - (C)