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Foram Desai 7 years, 9 months ago
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Sia ? 6 years, 5 months ago
Cost Price = ( SP * 100 ) / ( 100 + percentage profit).
Selling Price = (100 + Gain %) x C.P.
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Foram Desai 7 years, 9 months ago
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Sia ? 6 years, 6 months ago
L.H.S
= (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) = {tex}\frac { \sec A } { cosec ^ { 2 } A } - \frac { cosec A } { \sec ^ { 2 } A }{/tex} = sinA.tanA - cosA.cotA
Hence, Proved.
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Sia ? 6 years, 6 months ago
Given: The perpendicular from A on side BC of an{tex}\vartriangle {/tex} ABC intersect BC at Such that BC = 3CD
To prove:2AB2 = 2AC2 + BC2
Proof:In right triangle ADB,
AB2 = AD2 + BC2 (1).......[By Pythagoras theorem]
In right triangle ADC,
AC2 = AD2 + CD2 (2).....[By Pythagoras theorem]
Subtracting (2) from (1), we get
AB2 - AC2 = BD2 - CD2
= (BD + CD)(BD - CD)
= (BC)(3CD - CD) ........ {tex}\because {/tex} BD = 3CD(given)
= (BC)(2CD) = 2(BC)(CD)
=2(BC){tex}\left( {\frac{1}{4}BC} \right){/tex}
DB = 3CD
{tex} \Rightarrow \frac{{BD}}{{CD}} = 3{/tex}
{tex} \Rightarrow \frac{{DB}}{{CD}} + 1{/tex}
= 3 + 1
{tex} \Rightarrow \frac{{DB + CD}}{{CD}}=4{/tex}
{tex}\Rightarrow \frac{{BC}}{{CD}} = 4{/tex}
{tex}\Rightarrow CD = \frac{1}{4}BC = B{C^2}{/tex}
{tex}\Rightarrow {/tex} 2(AB2 - AC2) = BC2
{tex}\Rightarrow {/tex} 2AB2 - 2AC2 = BC2
{tex}\Rightarrow {/tex} 2AB2 = 2AC2 + BC2
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Retika Malhotra 7 years, 9 months ago
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