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Prove the following:`tan[pi/4+1/2cos^(-1)(a/b)]+tan[pi/4-1/2cos^(-1)(a/b)]=(2b)/a` |
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Answer» Let `1/2cos^-1(a/b) = x` Then, `a/b = cos2x->(1)` Now, `L.H.S. = tan(pi/4+1/2cos^-1(a/b))+tan(pi/4-1/2cos^-1(a/b))` `=tan(pi/4+x)+tan(pi/4-x)` `=(1+tanx)/(1-tanx)+(1-tanx)/(1+tanx)` `=((1+tanx)^2+(1-tanx)^2)/(1-tan^2x)` `=(1+tan^2x+2tanx+1+tan^2x-2tanx)/(1-tan^2x)` `=(2(1+tan^2x))/(1-tan^2x)` We know, `tan^2x = sec^2x-1` So, our expression becomes, `=(2sec^2x)/(2-sec^2x)` `=(2/cos^2x)/(2-1/cos^2x)` `=2/(2cos^2x-1)` `=2/(cos2x)` `=(2)/(a/b)` (From (1)) `=(2b)/a = R.H.S.` |
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