Loading [MathJax]/jax/output/HTML-CSS/jax.js


1)

limxπ/4tanx1cos2x is equal to 


A) 1

B) 0

C) -2

D) -1

Answer:

Option D

Explanation:

limxπ/4tanx1cos2x=limh0tan(π4+h)1cos2(π4+h)

[   x=π4+h]

=limh0(1+tanh1tanh)1cos(π2+2h)

=limh01+tanh1+tanhsin2h(1tanh)

= limh02tanh2sinhcosh(1tanh)

= limh01cos2h(1tanh)=1