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Question : 108 of 150
Marks:
+1,
-0
Solution:
(‌| 1+sin‌θ−cos‌θ |
| 1+sin‌θ+cos‌θ |
)2=λ(‌)⇒λ=‌[| (1+sin‌2θ+cos2θ+2sin‌θ−2sin‌θ‌cos‌θ−2‌cos‌θ)(1+cos‌θ) |
] |
| [(1+sin‌2θ+cos2θ+2sin‌θ+2sin‌θ‌cos‌θ+2‌cos‌θ)(1−cos‌θ)] |
⇒λ=‌[| 2(1+sin‌θ−cos‌θ−sin‌θ‌cos‌θ) | | (1+cos‌θ) |
] |
| [2(1+sin‌θ+cos‌θ+sin‌θ‌cos‌θ);(1−cos‌θ)] |
⇒λ=‌[| 1+sin‌θ−cos‌θ−sin‌θ‌cos‌θ+cos‌θ | | +sin‌θ‌cos‌θ−cos2θ−sin‌θcos2θ |
] |
| [1+sin‌θ+cos‌θ+sin‌θ‌cos‌θ−cos‌θ;−sin‌θ‌cos‌θ−cos2θ−sin‌θcos2θ] |
⇒λ=‌| 1+sin‌θ−cos2θ−sin‌θcos2θ |
| 1+sin‌θ−cos2θ−sin‌θcos2θ |
⇒λ=1
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