Question

PROVE THAT :
cot³Ф*sin³Ф÷(cosФ+sinФ)² + tan³Ф*cos³Ф÷(cosФ+sinФ)² = secФcosecФ-1÷cosecФ+secФ
plz ans this

Answer 1

Answer:

Step-by-step explanation:

LHS= cot³Фsin³Ф/(cosФ+sinФ)² + tan³Фcos³Ф/(cosФ + sinФ)²

cot³Ф=cos³Ф/sin³Ф

tan³Ф=sin³Ф/cos³Ф

=(cos³Ф/cosФ+sinФ )² + sin³Ф/(cosФ+sinФ)²

cos³Ф+sin³Ф/(cosФ+sinФ)²

we can write cos³Ф+sin³Ф as (cosФ+sinФ)(cos²Ф+sin²Ф-sinФcosФ) because a³+b³=a+b(a²+b²-ab)

so our equations becomes :

1-sinФcosФ/cosФ+sinФ

divide numerator and denominator by sinФcosФ

cosecФsecФ-1/cosecФ + secФ

= RHS

hence proved

hope for brainliest!!!!