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Ashish Kumar 6 years, 11 months ago

x common from R1 y common from R2 z common from R3 Then R1+R2+R3 ( Same as any example in ncert ) In example we have taken 3x common- hint
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Ashish Kumar 6 years, 11 months ago

Ncert two times + Ten year question papers can give you around 90 easily

Kashish Pushya 6 years, 11 months ago

Just Study Hard....For these two months
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Kashish Pushya 6 years, 11 months ago

Sec x ko 1/cosx likh do...and cosecx ko 1/sinx ...then take LCM
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Sanvi Sharma 6 years, 11 months ago

Any Modulus is not defined when its value is equal to 0 Ex:mod X is not derivable at 0.

Sanvi Sharma 6 years, 11 months ago

7/2
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Kashish Pushya 6 years, 11 months ago

1/sinxcosx As sin^2x+cos^2x=1 => sin^2x +cos^2x/sinxcosx =>Sin^2x/sinxcosx+cos^2x/sinxcosx => Sinx/cosx + cosx/sinx =>Tanx+cotx Now, integration... => /Tanx dx+/cotx dx => Sec^2x-cosec^2x
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Deeksha Singh 6 years, 11 months ago

Divide and multiple 2 in num and demo....than it will a formula 1/2$(2sin3xcosx)

Bharat Singh Rathore 6 years, 11 months ago

What
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Deeksha Singh 6 years, 11 months ago

(x-0)^2 +(y-0)^2 = (2)^2 ------ eq 1 for circle with its centre on origin and radius equal to 2 (x-2)^2 +y^2 = (2)^2 ------------eq 2 for 2nd circle whose center is (2,0) and radius again 2 Then equat both these equation and find the intersecting point Then shade the region enclosed by these circles and find the area with the limits provided using suitable method of finding area
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7-8
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Sunil Rajput 6 years, 11 months ago

I think. -1/6 or -8/15
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Sachin Singh 6 years, 11 months ago

Order not defined
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Prangana Phukon 6 years, 11 months ago

Since a, b and c are mutually perpendicular unit vectors, we get a.b=0, b.c=0, a.c=0 and |a|= |b|=|c|=1 Also,|2a+b+c|2=(2a+b+c)⋅(2a+b+c) =2a⋅(2a+b+c)+b⋅(2a+b+c)+c⋅(2a+b+c) =4 a⋅a+2 a⋅b+2 a⋅c+2 b⋅a+ b⋅b+b⋅c+2 c⋅a+c⋅b+c⋅c a⋅a=|a|^2, b⋅b=|b|^2, c⋅c=|c|^2, and we know that dot product is commutative. So , |2a+b+c|2=4 |a|2+2a⋅b+2 a⋅c+2 a⋅b+|b|2+b⋅c+2a⋅c+b⋅c+|c|2 =4 |a|2+|b|2+|c|2+4 a⋅b+4 a⋅c+2 b⋅c =4(1)+1+1+4(0)+4 (0)c+2 (0) =6This further implies that, |2a+b+c| = √6

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