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Ask QuestionPosted by Piyush Bothra 6 years, 7 months ago
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Posted by Harpreet Singh 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
tan(A - B) = [tan(A) - tan(B)]/[1 + tan(A)tan(B)].
Thus:
tan(13π/12)
= tan(4π/3 - π/4), from the hint
= [tan(4π/3) - tan(π/4)]/[1 + tan(4π/3)tan(π/4)], from the above formula
= (-1 + √3)/(1 + √3)
= -(1 - √3)/(1 + √3)
= -(1 - √3)^2/[(1 + √3)(1 - √3)], by rationalizing
= -(1 - √3)^2/(1 - 3), via difference of squares
= (1/2)(1 - √3)^2
= (1/2)(1 - 2√3 + 3)
= (1/2)(4 - 2√3)
= 2 - √3.
Posted by Harpreet Singh 6 years, 7 months ago
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Posted by Kshitiz Chaudhary 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
since,sin60=√ 3/2
= √ 3/2( sin20sin40sin80)
=√ 3/2( sin20sin80sin40)
=√ 3/4 [(2sin20sin40)sin80]
on applying [cos(A-B)-cos(A+B) = 2sinAsinB]
we get,
= √ 3/4 (cos20-cos60)sin80 [since,cos(-a)=cosa]
= √ 3/4(cos20sin80-cos60sin80)
= √ 3/8(2sin80cos20-sin80)
= √ 3/8(sin100+sin60-sin80)
= √ 3/8( √ 3/2+sin100-sin80 )
= √ 3/8( √ 3/2+sin(180-80)-sin80 )
= √ 3/8( √ 3/2+sin80-sin80 ) [since,sin(180-a)=sina]
= √ 3/8( √ 3/2)
= 3/16
Posted by Harpreet Singh 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
Posted by Yash Goswami????? 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
Let P(n) = {tex}1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + .... + n(n + 1)(n + 2){/tex}{tex} = \frac{{n(n + 1)(n + 2)(n + 3)}}{4}{/tex}
For n = 1
{tex}P(1) = 1 \times 2 \times 3 = \frac{{1 \times 2 \times 3 \times 4}}{4} \Rightarrow 6 = 6{/tex}
{tex}\therefore {/tex} P (1) is true
Let P(n) be true for n = k.
{tex}\therefore P(k) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ...{/tex}{tex} + (k + 1)(k + 2) = \frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex} ... (i)
For n = k + 1
{tex}P(k + 1) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + {/tex}{tex}k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3){/tex}
{tex} = \frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex} + (k + 1) (k + 2) (k + 3) [Using (i)]
{tex} = (k + 1)(k + 2)(k + 3)\left[ {\frac{{k + 4}}{4}} \right]{/tex}
{tex} = \frac{{(k + 1)(k + 2)(k + 3)(k + 4)}}{4}{/tex}
{tex}\therefore {/tex} P(k + 1) is true.
Thus P(k) is true {tex} \Rightarrow {/tex} P (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all {tex}n \in N{/tex}.
Posted by Surendra Jangir 6 years, 7 months ago
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Posted by Aniketa Kumari 6 years, 7 months ago
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Sia ? 6 years, 7 months ago
The given system of equations is
{tex}3x - \frac{{y + 7}}{{11}} + 2 = 10{/tex} ...(i)
{tex}2y + \frac{{x - 11}}{7} = 10{/tex} ....(ii)
From (i), we get
{tex}\frac{{33x - y - 7 + 22}}{{11}} = 10{/tex}
{tex}\Rightarrow{/tex} {tex}33x - y + 15 = 10 × 11{/tex}
{tex}\Rightarrow{/tex} {tex}33x + 15 - 110 = y{/tex}
{tex}\Rightarrow{/tex} {tex}y = 33x - 95{/tex}
From (ii), we get
{tex}\frac{{14y + x + 11}}{7} = 10{/tex}
{tex}\Rightarrow{/tex} {tex}14y + x + 11 = 10 × 7{/tex}
{tex}\Rightarrow{/tex} {tex}14y + x + 11 = 70{/tex}
{tex}\Rightarrow{/tex} {tex}14y + x = 70 - 11{/tex}
{tex}\Rightarrow{/tex} {tex}14y + x = 59{/tex} ....(iii)
Substituting y = 33x - 95 in (iii), we get
14(33x - 95) + x = 59
{tex}\Rightarrow{/tex} 462x - 1330 + x = 59
{tex}\Rightarrow{/tex} 463x = 59 + 1330
{tex}\Rightarrow{/tex} 463x = 1389
{tex}\Rightarrow x = \frac{{1389}}{{463}} = 3{/tex}
Putting x = 3, in y = 33x - 95, we get
y = 33 {tex}\times{/tex} 3 - 95
{tex}\Rightarrow{/tex} y = 99 - 95 = 4
{tex}\Rightarrow{/tex} y = 4
Hence, Solution of the given system of equation is x = 3, y = 4.
Posted by Kavya Yadav 6 years, 7 months ago
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Yogita Ingle 6 years, 7 months ago
We know that 180° = π radian.
40° 20' = 40 1/3 degree = π /180 × 121/3 radian = 121π /540radian
Therefore 40° 20′ = 121π/540 radian.
Posted by __@___ Goel 6 years, 7 months ago
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Anuska Thakur 6 years, 7 months ago
Yogita Ingle 6 years, 7 months ago
Hence,
= 343° + 38′ + 10.9″ = 343°38′ 11″ approximately.
Hence 6 radians = 343° 38′ 11″ approximately.</div>
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