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Prove:- (tan A + cosec B) square - (cot B - sec A)square = 2tan A cot B (cosec A + sec B) |
| Answer» 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 secARearranging above equation we getcosec²B - 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 (\xa0cosA/sinA) }= 2tanA cotB ( 1/cosB + 1/sinA) = 2tanA cotB (secB + cosecA)= 2tanA cotB (cosecA + secB)Read more on Brainly.in - https://brainly.in/question/2314597#readmore | |