Prove that :-
(cot³A.sin³A)/(cosA+sinA)² + (tan³A.cos³A)/(cosA+sinA)² = (secA.cosecA-1)/(cosecA+secA)
need the answer ASAP
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Step-by-step explanation:
LHS
We have formula ,
a³ +b³ = a+b(a² -ab+b²)
Important steps -
(a+b)² = (a+b)(a+b) (step 3)
.
sin ²A +cos ²A = 1 (step 3)
tan A = sin A/ cos A ( used in step 1)
cot A = cos A / sin A
RHS - Steps
Cosec A = 1/sin A
Sec A = 1/ cos A
Hence proved
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