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| 1. |
if tan a 3/4 then prove that sinA cosA 12/25 |
| Answer» Given, tan A =\xa0{tex}\\frac { 3 } { 4 }{/tex}⇒ tan A =\xa0{tex}\\frac { P } { B } = \\frac { 3 x } { 4 x }{/tex}\xa0[ from figure ]H2\xa0= P2\xa0+ B2\xa0[By Pythagoras theorem]= (3x)2\xa0+ (4x)2= 9x2\xa0+ 16x2⇒ H2\xa0= 25x2or, H2=(5x)2⇒ H = 5xTherefore,sin A =\xa0{tex}\\frac { P } { H } = \\frac { 3 x } { 5 x } = \\frac { 3 } { 5 }{/tex}and cos A =\xa0{tex}\\frac { B } { H } = \\frac { 4 x } { 5 x } = \\frac { 4 } { 5 }{/tex}Now, LHS = sin A cos A\xa0{tex}= \\frac { 3 } { 5 } \\times \\frac { 4 } { 5 } = \\frac { 12 } { 25 }{/tex} = RHSHence, proved. | |