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Posted by Anja8 Rawat 5 years, 1 month ago
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Kunal Jadhav 5 years, 1 month ago
Posted by Saranya Rajesh Kumar Saranya 5 years, 1 month ago
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Posted by Babita Pathak 5 years, 1 month ago
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Sehaj Gamer? 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
P = Rs. 800 [Amount deposited]
T = 3 years
R = 12%
S.I. = PRT/100
S.I. = 800 × 3 × 12/100
S.I. = 8 × 3 × 12
S.I. = 288
Amount recieved = Rs. 288 + Rs. 800 {Principal + Simple Interest}
Rs. 1088
Posted by Samarth Kapdi 5 years, 1 month ago
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Posted by Harish Baghel 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
The series of odd integers from 1 to 2001 is,1,3,5,7..............,2001
a=1
d=3−1=2
l=2001
an = a+(n−1)d
2001=1+(n−1)2
2000=2(n−1)
∴n=1001
Sn=n/2 [a+l]
=1001/2 [1+2001]
=10,02,001
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Gaurav Seth 5 years, 1 month ago
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain r2 (cos2 θ + sin2 θ) = 1+ 1
⇒ r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
This is the required polar form.
Posted by Rohit Kumar 5 years, 1 month ago
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Posted by Manali Mali 5 years, 1 month ago
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Aadya Singh 5 years, 1 month ago
=5(t2- 2t-3 )
=5 (t2 - 3t + t - 3)
=5 [t(t-3) +1(t-3)]
= 5 (t+1) (t-3) ?
Posted by Santosh Chaudhary 5 years, 1 month ago
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Posted by Pragati Goyal 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Prove the following identites
sec4x - sec2x = tan4x + tan2x
SOLUTION
LHS = sec4x - sec2x
= sec2x(sec2x - 1)
= (tan2x + 1)(tan2x) (∵ sec2x - tan2x = 1)
= tan4x + tan2x
=RHS
Hence proved.
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Sehaj Gamer? 5 years, 1 month ago
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