A) 2
B) 1
C) -2
D) -1
Correct Answer: C
Solution :
\[\frac{df\,\left( \tan \,x \right)}{dg\,\left( \sec \,x \right)}=\frac{f'\left( \tan \,x \right)\,{{\sec }^{2}}\,x}{g'\left( \sec \,x \right)\sec \,x\,\tan \,x}\] \[=\frac{f'\left( \tan \,x \right)\,\sec \,x}{g'\,\left( \sec \,x \right)\,\tan \,x}\] \[\therefore \,{{\left[ \frac{df\left( \tan \,x \right)}{dg\,\left( \sec \,x \right)} \right]}_{x=\pi /4}}=\frac{f'\left( 1 \right)\sqrt{2}}{g'\left( 2 \right).1}=\frac{2\sqrt{2}}{4\,.\,1}=\frac{1}{\sqrt{2}}\].
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