InterviewSolution
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InterviewSolution
| 1. |
(1+cotA+tan) ( sinA-cisA) = sinAtanA -cot A cosA |
| Answer» L.H.S\xa0= (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA........(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} - \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} - \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} - \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} - \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A - \\cos A \\times \\cot A{/tex}.......(2)From (1) & (2),(1 + cotA + tanA) (sinA - cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } - \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA - cosA.cotA\xa0Hence, Proved. | |