Question

how to solve this question
need easy explanations

Answer 1

tan²x=1- a²      i.e.       sec²x =2-a²...(1)

LHS=secx + tan³x cosecx = 1/cosx + sin²x/cos³x= 1/cosx + (1-cos²x )/cos³x

= 1/cosx + 1/cos³x – 1/cosx= 1/cos³x= sec³x= (2-a²)³/² from eqs 1

Answer 2

well first find out tan0
then find out sec0 using trigo formulas
similarly find cos0 using sec0
using cos0 find out sin0 and take its reciprocal
now substitue in the above LHS
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