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`tan(pi/4+1/2cos^-1x)+tan(pi/4-1/2cos^-1x)`, `x!=0` is equal to |
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Answer» Let `1/2 cos^-1x = theta=> cos2theta = x` Then, our expression becomes, `tan(pi/4+theta)+tan(pi/4-theta)` `=(1+tantheta)/(1-tantheta) + (1-tantheta)/(1+tantheta)` `(1+tan^2theta+2tantheta+1+tan^2theta-2tantheta)/(1-tan^2theta)` `=2((1+tan^2theta)/(1-tan^2theta))` As `cos2theta = (1-tan^2theta)/(1+tan^2theta)` So, our expression becomes, `=2/(cos 2theta) = 2/x`, which is the desired value for our expression. |
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