if cot thita +cos thita=p and cot thita=q then the value of p²-q² is:
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Answer with Step-by-step explanation:
cotФ + cosФ = p ------(1)
cotФ - cosФ = q ------(2)
LHS = P² - q²
= (cotФ + cosФ)² - (cotФ - cosФ)²
= cot²Ф + cos²Ф + 2cosФ.cotФ - cot²Ф - cos²Ф + 2cosФ.cotФ
= 4cotФ.cosФ
RHS = 4√pq
= 4√{(cotФ + cosФ)(cotФ - cosФ)}
= 4√{cot²Ф - cos²Ф}
= 4√{1/tan²Ф - 1/sec²Ф}
= 4√{(sec²Ф - tan²Ф)/tan²Ф.sec²Ф}
= 4√{1/tan²Ф.sec²Ф} [∵ from Identity sec²x -tan²x = 1 ]
= 4√{cot²Ф.cos²Ф}
= 4cotФ.cosФ
LHS = RHS [hence proved]
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