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`2"tan"(tan^(-1)(x)+tan^(-1)(x^3)),w h e r ex in R-{-1,1},`is equal to`(2x)/(1-x^2)``t(2tan^(-1)x)``tan(cot^(-1)(-x)-cot^(-1)(x))``"tan"(2cot^(-1)x)` |
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Answer» Let `tan^-1x = alpha => tanalpha = x` Let `tan^-1x^3 = beta => tan beta = x^3` Then, `2tan(tan^-1x+tan^-1x^3) = 2tan(alpha+beta)` `=2[(tanalpha+tanbeta)/(1-tanalphatanbeta)]` `=2[(x+x^3)/(1-x(x^3))]` `=2[(x(1+x^2))/(1-x^4)]` `=2[(x(1+x^2))/((1-x^2)(1+x^2))]` `=(2x)/(1-x^2)->(1)` `=(2tanalpha)/(1-tan^2alpha)` `=tan2alpha` `=tan(2tan^-1x)->(2)` `=tan(2(pi/2-cot^-1x)` `=tan(pi-2cot^-1x)` `=tan(cot^-1(-x)-cot^-1(x))->(3)` From (1),(2), (3), options, `a`,`b` and `c` are correct options. |
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