value of tanx.

(a+2)sinx + (2a-1)cosx = (2a +1)

(a+2)2sinx/2cosx/2 = 2a(1-cosx) + (1+cosx)

(a+2)2sinx/2cosx/2 = 4asin²x/2 + 2cos²x/2

divide both sides by 2 and then multiply both sides by sec²x/2 [we can do the multiplication because we can see from the equation that x is not equal to Ï€(pi)]

tanx/2(a+2) =2atan²x/2 + 1
this is a quadratic eqtn in tanx/2....solve it

the answer will be (b) and (d) both....

Let \alpha be such that

\tan \alpha = \frac{2a-1}{a+2}

Then the given equation may be written as

\sin (x+\alpha) = \frac{2a+1}{\sqrt{5(a^2+1)}}

So that

\tan (x+\alpha) = \frac{2a+1}{a-2}

Hence \tan x = \tan (x+\alpha-\alpha) = \frac{\tan(x+\alpha) - \tan \alpha}{1+\tan (x+\alpha) \tan \alpha} = \frac{2a}{a^2-1}

we can write sinx =
2tan(x/2)1+tan²(x/2)
similalry, cosx =
(1-tan²(x/2))(1+tan²(x/2))
we can solve for tan(x/2)...
i hope i am correct.