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Question : 63 of 160
Marks:
+1,
-0
Solution:
Given,
xsin‌(α+y)=sin‌y and
y=‌,
xsin‌(α+y)=sin‌y ‌⇒‌‌‌=‌‌⇒‌‌‌| sin‌α‌cos‌y+cos‌α‌s‌i‌n‌y |
| sin‌y |
=‌‌⇒‌‌sin‌α‌cot‌y+cos‌α=‌‌⇒‌‌cot‌y=‌| 1−x‌cos‌α |
| xsin‌α |
‌⇒‌‌tan‌y=‌| xsin‌α |
| 1−x‌cos‌α |
‌⇒‌‌y=tan−1[‌| xsin‌α |
| 1−x‌cos‌α |
] y′=‌‌| 1 |
1+(‌| xsin‌α | | 1−x‌cos‌α |
)2 |
‌[‌| (1−x‌cos‌α)sin‌α−xsin‌α(−cos‌α) |
| (1−x‌cos‌α)2 |
] ⇒y′=‌‌| (1−x‌cos‌α)2 |
| (1−x‌cos‌α)2+(xsin‌α)2 |
‌[‌| sin‌α−x‌cos‌α‌s‌i‌n‌α+x‌cos‌α‌s‌i‌n‌α |
| (1−x‌cos‌α)2 |
] ‌⇒y′=‌| sin‌α |
| 1+x2cos2α−2x‌cos‌α+x2sin‌2α |
‌⇒y′=‌| sin‌α |
| x2−2x‌cos‌α+1 |
Here,
m=sin‌α and
n=−cos‌αm2‌=sin‌2α‌=1−cos2α=1−n2
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