Question
Prove : $\left(1+\cot ^2 \theta\right)(1+\cos \theta)(1-\cos \theta)=1$
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Answer
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Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity :- Natural numbers between 1 to 140 divisible by 4 are,
4,8,12,16,......136
Here d=4, therefore this sequence is an A.P.
a=4, d=4,$t _{ n }=136, \quad S _{ n }=?$
$t_n=a+(n-1) d$
_____= 4+(n-1)×4
_____= (n-1)×4
n =_____
Now,
$S _{ n }=\frac{ n }{2}+\left[ a + t _{ n }\right]$
$S_n=17 \times$____
$S _{ n }=$____
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is _____
→Activity :- Natural numbers between 1 to 140 divisible by 4 are,
4,8,12,16,......136
Here d=4, therefore this sequence is an A.P.
a=4, d=4,$t _{ n }=136, \quad S _{ n }=?$
$t_n=a+(n-1) d$
_____= 4+(n-1)×4
_____= (n-1)×4
n =_____
Now,
$S _{ n }=\frac{ n }{2}+\left[ a + t _{ n }\right]$
$S_n=17 \times$____
$S _{ n }=$____
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is _____
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial such that $\alpha+\beta=24$ and $\alpha-\beta=8,$ find a quadratic polynomial have $\alpha$ and $\beta$ as its zeroes.
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