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Ask QuestionPosted by Abhishek Shekhawat 8 years, 4 months ago
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Posted by Sam Anugraha Tirkey 8 years, 4 months ago
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Posted by Drakshi Chopra 8 years, 4 months ago
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Priya Singh 7 years, 11 months ago
=(1/8)(cos80°+cos100°)+(1/8)×(1/2)
=(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)
=(1/8)(2cos90°cos10°)+(1/16)
=0+(1/16) [cos90°=0]
=1/16 (proved)
Posted by Nisha Seth 8 years, 4 months ago
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Posted by Tanmay Ganguly 8 years, 4 months ago
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Vidhi Jain 8 years, 4 months ago
Posted by Mukul Gaur 8 years, 4 months ago
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Posted by Ashitosh Jadhav 8 years, 4 months ago
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Posted by Tehzeeb Alam 8 years, 4 months ago
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Posted by Abhay Singh 8 years, 4 months ago
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Dharmendra Kumar 8 years, 4 months ago
x2+(p+2)x+2p=0
On comparing with ax2+bx+c=0 we get
a=1 b=p+2 c=2p
Now since roots of the given quadratic equation are distinct integer.
Therefore, Discriminent D=b2-4ac > 0
(p+2)2-4×1×2p > 0
p2+2×p×2+22 -8p > 0
p2+4p+4-8p > 0
p2 -4p +4 > 0
p2-2p-2p+4 > 0
p(p-2)-2(p-2) > 0
(p-2)(p-2) > 0
(p-2)2 > 0
p-2 > 0
p > 2
So roots of the given quadratic eq. is distinct integer if p>2.
Posted by Khushbu Vohra 8 years, 4 months ago
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Posted by Khushbu Vohra 8 years, 4 months ago
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Posted by Saloni Kumari 8 years, 4 months ago
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Posted by Tanishque Kumar 8 years, 4 months ago
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Posted by Vishal Sharma 8 years, 4 months ago
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Posted by Jagdish Verma 8 years, 4 months ago
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Dharmendra Kumar 8 years, 4 months ago
We can find its value as follows:-
tan15°= tan(45°-30°)
= {tex}{tan 45° - tan30°\over 1+tan45°tan30°}{/tex}
= {tex}{1- {1\over {\sqrt 3}}}\over1+1×{1\over {\sqrt 3}}{/tex}
= {tex}{{{\sqrt 3}-1}\over {\sqrt 3}}\over {{{\sqrt 3}+1}\over {\sqrt 3}}{/tex}
= {tex}{\sqrt 3}-1\over {\sqrt 3}+1{/tex}
Posted by Vidhi Jain 8 years, 4 months ago
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