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Posted by Shivam Kumar 6 years, 1 month ago
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Yogita Ingle 6 years, 1 month ago
We need to find the H.C.F. of 963 and 657 and express it as a linear combination of 963 and 657. By applying Euclid’s division lemma, 963 = 657 x 1 + 306.
Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306
657 = 306 x 2 + 45.
Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 45
306 = 45 x 6 + 36.
Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36
45 = 36 x 1 + 9.
Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9
36 = 9 x 4 + 0.
Therefore, H.C.F. = 9.
Now, 9 = 45 – 36 x 1
= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6
= 45 x 7 – 306 x 1 = [657 -306 x 2] x 7 – 306 x 1
= 657 x 7 – 306 x 14 – 306 x 1
= 657 x 7 – 306 x 15
= 657 x 7 – [963 – 657 x 1] x 15
= 657 x 7 – 963 x 15 + 657 x 15
= 657 x 22 – 963 x 15.
Hence, obtained.
Hence x = -15 and y = 22
Posted by Aryan Saxena 6 years, 1 month ago
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Yogita Ingle 6 years, 1 month ago
Given that Sin A + 2 cos A = 1
Squaring on both sides, we get
(sin A + 2 cos A)2 = 1
We know that (a+b)2 = a2 + b2 + 2ab.
(sin2 A + 4 cos2 A + 4 sin A cos A) = 1
4 cos2 A + 4 sin A cos A = 1 - sin2 A
4 cos2 A + 4 sin A cos A = cos2 A
3 cos2 A + 4 sin A cos A = 0
3 cos2 A = - 4 sin A cos A ---- (1).
Given 2 sin A - cos A
Squaring on both sides, we get
(2 sin A - cos A)2 = 4 sin2 A + cos2 A - 4 sin A cos A
= 4 sin2 A + cos2 A + 3 cos2 A
= 4 sin2 A + 4 cos2 A
= 4(sin2 A + cos2 A)
= 4.
2 sin A - cos A = 2.
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Dishant Rajput 6 years, 1 month ago
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