if (a+2)sinx + (2a-1)cosx = (2a+1) find tanx.
the options were
(a) 3/4
(B) 4/3
(C) 2a/(a2+1)
(D) 2a/(a2-1)
well we can see that option B holds.
i want the solution.
1(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 π(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....
341Let \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}
3we 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.
