JEE Math Practice Question
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 :
- A.$${1 \over {3\sqrt 2 }}$$
- B.$${1 \over {6\sqrt 2 }}$$
- C.$${3 \over {2\sqrt 2 }}$$
- D.$${1 \over 6}$$
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