STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

${(m + 2)^2}\,{t^2} + 2(m + 2)\,(2m - 1)t + {(2m - 1)^2} = {(2m + 1)^2}\,(1 + {t^2})$

$ \Rightarrow \,3\,(1 - {m^2})\,{t^2} + (4{m^2} + 6m - 4)\,t - 8m = 0$

$ \Rightarrow \,(3t - 4)\,[(1 - {m^2})\,t + 2m = 0$,

जो कि सत्य है यदि $t = \tan \theta  = \frac{4}{3}$ या $\tan \theta  = \frac{{2m}}{{{m^2} - 1}}$.