Prove that tan cube theta divided by one plus tan squared theta plus Cot cube theta divided by one plus Cot Square theta is equals two sec theta into course sec theta -2 sine theta cos theta.
Answers
Answer:
Step-by-step explanation:
To proved , [tan³∅/(1 +tan²∅)]+[cot³∅(1+cot²∅)]=sec∅cosec∅ - 2sin∅2cos∅
LHS:-
[tan³∅/(1+tan²∅)]+[cot³∅/(1+cot²∅)] = [tan³∅/sec²∅]+[cot³∅/cosec²∅]
= [(sin³∅/cos³∅)/(1/cos²∅)]+[(cos³∅/sin³∅)/[1/sin²∅]
=[sin³∅/cos∅]+[cos³∅/sin∅]
=[sin⁴∅ + cos⁴∅]/[sin∅ cos∅]
=[(sin²∅+cos²∅)²-2 sin²∅ cos²∅]/[sin∅ cos∅]
=[1²-2 sin²∅ cos²∅]/[sin∅ cos∅]
=[1/sin∅ cos∅] - [2 sin²∅ cos²∅/sin∅ cos∅]
=sec∅ cosec∅ - 2 sin∅ cos∅
Therefore LHS = RHS
Step-by-step explanation:
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