If x $=$ 3 tan t and y $=$ 3 sec t, then the value of ${{{d^2}y} \over {d{x^2}}}$ at t $ = {\pi \over 4},$ is :

Question

${1 \over {3\sqrt 2 }}$

${1 \over {6\sqrt 2 }}$

${3 \over {2\sqrt 2 }}$

${1 \over 6}$

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