For $$y = {\sin ^{ - 1}}\left\{ {{{5x + 12\sqrt {1 - {x^2}} } \over {13}}} \right\};\left| x \right| \le 1$$, if $$a(1 - {x^2}){y_2} + bx{y_1} = 0$$ then (a, b) =

If $$0 \le A \le {\pi \over 4}$$, then $${\tan ^{ - 1}}\left( {{1 \over 2}\tan 2A} \right) + {\tan ^{ - 1}}(\cot A) + {\tan ^{ - 1}}({\cot ^3}A)$$

The possible values of x, which satisfy the trigonometric equation

$${\tan ^{ - 1}}\left( {{{x - 1} \over {x - 2}}} \right) + {\tan ^{ - 1}}\left( {{{x + 1} \over {x + 2}}} \right) = {\pi \over 4}$$ are

If $$f(x) = {\tan ^{ - 1}}\left[ {{{\log \left( {{e \over {{x^2}}}} \right)} \over {\log (e{x^2})}}} \right] + {\tan ^{ - 1}}\left[ {{{3 + 2\log x} \over {1 - 6\log x}}} \right]$$, then the value of f''(x) is equal to