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Question : 22 of 160
Marks:
+1,
-0
Solution:
‌(‌)2−(‌)2=a‌cos‌b‌θ‌⇒(‌| 3sin‌θ−4sin‌3θ |
| sin‌θ |
)−(‌| 4cos3θ−3‌cos‌θ |
| cos‌θ |
)=a‌cos‌b‌θ‌⇒(3−4sin‌2θ)2−(4cos2θ−3)2=a‌cos‌b‌θ‌⇒9+16sin‌4θ−24sin‌2θ−16cos4θ−9+24cos2θ=a‌cos‌b‌θ‌⇒24(cos2θ−sin‌2θ)−16(cos4θ−sin‌4θ)=acosbθ‌⇒8(cos2θ−sin‌2θ)(3−2(cos2θ+sin‌2θ))=a‌cos‌b‌θ,‌⇒8‌cos‌2‌θ(3−2)=a‌cos‌b‌θ‌⇒8‌cos‌2‌θ=a‌cos‌b‌θ⇒a=8,b=2‌∴a:b=8:2=4:1
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