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| 1. |
To prove (1+tanAtanB)²+(tanAtanB)²=sec²Asec²B |
| Answer» We have,L.H.S = (1 + tanA tanB)2 + (tan A - tanB)2{tex}\\Rightarrow{/tex}\xa0L.H.S = (1 + 2tanA.tanB + tan2A.tan2B) + (tan2A + tan2B-2tanA.tanB) {tex}\\left[\\because\\left(a+b\\right)^2=a^2+2ab+b^2,\\;\\left(a-b\\right)^2=a^2-2ab+b^2\\right]{/tex}{tex}\\Rightarrow{/tex}\xa0L.H.S = 1 + tan2A.tan2B + tan2A + tan2B{tex}\\Rightarrow{/tex}\xa0L.H.S = (1 + tan2A) + (tan2B + tan2A.tan2B){tex}\\Rightarrow{/tex}\xa0L.H.S = (1 + tan2A) + tan2B(1 + tan2A) {tex}\\left[\\;taking\\;\\tan^2B\\;as\\;a\\;common\\;\\right]{/tex}{tex}\\Rightarrow{/tex} L.H.S = (1 + tan2A)(1 + tan2B) = sec2A sec2B = R.H.S {tex}\\left[\\because1+\\tan^2\\theta\\;=\\;sec^2\\theta\\right]{/tex}therefore,(1 + tanA.tanB)2 + (tan A - tanB)2\xa0{tex}={/tex}\xa0sec2A.sec2BHence proved. | |