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Sia ? 6 years, 5 months ago
We have, {tex}asin\theta+bcos\theta=c{/tex}
On squaring both sides, we get
{tex}(asin\theta+bcos\theta)^2=c^2{/tex}
(a sin θ)2 + (b cos θ)2 + 2(a sin θ) (b cos θ) = c2
⇒ a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ = c2
⇒ a2(1 – cos2 θ) + b2 (1 – sin2 θ) + 2 ab sin θ cos θ = c2 {tex}[\because sin^2\theta+cos^2\theta=1]{/tex}
⇒ a2 – a2 cos2 θ + b2 – b2 sin2 θ + 2ab sin θ cos θ = c2
⇒ –a2 cos2 θ – b2 sin2 θ + 2ab sin θ cos θ = c2 – a2 – b2
Taking Negative common,
⇒ a2 cos2 θ + b2 sin2 θ – 2ab sin θ cos θ = a2 + b2 – c2
⇒ (a cos θ)2 + (b sin θ)2 – 2(a cos θ) (b sin θ) = a2 + b2 – c2
⇒ {tex}(acos\theta-bsin\theta)^2=a^2+b^2-c^2{/tex}
⇒{tex}acos\theta-bsin\theta{/tex} = {tex}\pm \sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}
Hence proved, {tex}acos\theta-bsin\theta{/tex} = {tex}\sqrt{a^2+b^2-c^2}{/tex}
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The existing Mathematics examination is the Standard Level Examination.
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Sia ? 6 years, 5 months ago
Let a be the first term and d be the common difference of the given AP. Then,
Sm = sum of first m terms of the given AP;
Sn = sum of first n terms of the given AP.
{tex}\frac { S _ { m } } { S _ { n } } = \frac { m ^ { 2 } } { n ^ { 2 } } \Rightarrow \frac { \frac { m } { 2 } [ 2 a + ( m - 1 ) d ] } { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } = \frac { m ^ { 2 } } { n ^ { 2 } }{/tex}
{tex} \Rightarrow \frac { 2 a + ( m - 1 ) d } { 2 a + ( n - 1 ) d } = \frac { m } { n }{/tex}
{tex}\Rightarrow{/tex} 2an+mnd-nd= 2am+mnd-md
{tex}\Rightarrow{/tex} 2an-2am=nd-md
{tex}\Rightarrow{/tex} 2a(n - m) = d(n - m) {tex}\Rightarrow{/tex}2a=d...(i)
{tex}\therefore \quad \frac { T _ { m } } { T _ { n } } = \frac { a + ( m - 1 ) d } { a + ( n - 1 ) d } = \frac { a + ( m - 1 ) \cdot 2 a } { a + ( n - 1 ) \cdot 2 a }{/tex} [from (i)]
{tex}= \frac { a + 2 a m - 2 a } { a + 2 a n - 2 a } = \frac { 2 a m - a } { 2 a n - a } = \frac { a ( 2 m - 1 ) } { a ( 2 n - 1 ) } = \frac { 2 m - 1 } { 2 n - 1 }{/tex}.
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Sia ? 6 years, 5 months ago
Let O be the common centre of the two circles
and AB be the chord of the larger circle which touches the smaller circle at C.
Join OA and OC.
Then, OA = a and OC = b.
Now, {tex} \mathrm { OC } \perp A B{/tex} and OC bisects AB [{tex}\because{/tex} the chord of the larger circle touching the smaller circle, is bisected at the point of contact].
In right {tex}\triangle A C O{/tex}, we have
OA2 = OC2+AC2 [by Pythagoras' theorem]
{tex}\Rightarrow{/tex}AC = {tex}\sqrt { O A ^ { 2 } - O C ^ { 2 } } = \sqrt { a ^ { 2 } - b ^ { 2 } }{/tex}.
{tex}\therefore{/tex} AB = 2AC = {tex}2 \sqrt { a ^ { 2 } - b ^ { 2 } }{/tex} {tex}[ \because \text { C is the midpoint of } A B ]{/tex}
i.e, required length of the chord AB = {tex}2 \sqrt { a ^ { 2 } - b ^ { 2 } }{/tex}
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