Question

prove that cosec²ø×tan²ø-1=tan²ø
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Answer 1

Step-by-step explanation:

tanA + cosecB)² - (cotB - secA)² = tan²A + cosec²B + 2 tanA cosecB -(cot²B + sec²A - 2 cotB secA) ( using (a±b)² = a² + b² ± 2ab) = tan²A + cosec²B + 2tanA cosecB - cot²B - sec²A + 2cotB secA Rearranging above equation we get cosec²B - cot²B - sec²A + tan²A + 2tanA cosecB + 2cotB secA = (cosec²B - cot²B) - (sec²A - tan²A) + 2tanA cotB ( cosecB/cotB + secA/tanA ) = 1 - 1 + 2tanA cotB { (1/sinB)/(cosB/sinB) + (1/cosA)/(sinA/cosA) } ( As cosec²ø - cot²ø = 1 sec²ø - tan²ø = 1) = 0 + 2tanA cotB { (1/sinB) x (sinB/cosB) + (1/cos A) x ( cosA/sinA) } = 2tanA cotB ( 1/cosB + 1/sinA) = 2tanA cotB (secB + cosecA) = 2tanA cotB (cosecA + secB)

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