Question

Prove that tan⁴x = tan²x sec²x - sec²x +1

Answer 1

Step-by-step explanation:

$rhs = { \tan( \alpha ) }^{2} { \sec( \alpha ) }^{2} - { \sec( \alpha ) }^{2} + 1 \\ \\ = { \tan( \alpha ) }^{2} { \sec( \alpha ) }^{2} - ( { \sec( \alpha ) }^{2} - 1) \\ \\ = { \tan( \alpha ) }^{2} { \sec( \alpha ) }^{2} - { \tan( \alpha ) }^{2} \\ \\ = { \tan( \alpha ) }^{2} ( { \sec( \alpha ) }^{2} - 1) \\ \\ = { \tan( \alpha ) }^{2} . { \tan( \alpha ) }^{2} \\ \\ = { \tan( \alpha ) }^{4} \\ \\ = lhs.$

Answer 2

the answer is

taking LHS

= tan⁴x

=(tan²x) (tan²x)

=(tan²x) (sec²x-1). [by using 1+tan²x =sec²x]

=tan²x sec²x - tan²x

=tan²x sec²x - (sec²x-1). [by using 1+ tan²x=sec²x]

=tan²x sec²x - sec²x + 1

HENCE, Lhs=Rhs