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Question : 66 of 80
Marks:
+1,
-0
Solution:
Here,
x‌sin(a+y)+sin‌a‌cos(a+y)=0 ....(i)
On differentiating w.r.t.
x, we get
[x‌sin(a+y)] + [sin‌a‌cos(a+y)]=0 ⇒ [x {sin(a+y)}+sin(a+y)‌ ‌(x)‌ ] +sin‌a‌ ‌‌cos(a+y)=0 (using product rule and chain rule)
⇒
[x‌cos(a+y)‌ ‌(a+y)+sin(a+y)‌ ] +sin‌a‌ [−sin(a+y)‌ ‌(a+y)‌ ]=0 ⇒
[x‌cos(a+y)‌ (0+ )+sin(a+y)‌ ] −sin‌a‌sin(a+y)‌ (0+ )=0 ⇒
x‌cos(a+y)‌ ‌+sin(a+y) −sin‌a‌sin(a+y)‌ ‌=0 ⇒
[x‌cos(a+y)−sin‌a‌sin(a+y)] =−sin(a+y) from Eq. (i), putting
x=−sin‌a‌ ‌ ⇒
[−sin‌a‌ ‌−sin‌a‌sin(a+y)‌ ] =−sin(a+y) ⇒
− [ | sin‌a‌cos2(a+y)+sin‌a‌sin2(a+y) |
| sin(a+y) |
] =−sin(a+y) ⇒
=sin(a+y) [ | sin(a+y) |
| sin‌a‌ {cos2(a+y)+sin2(a+y) } |
] ⇒
= ‌‌‌(∵sin2θ+cos2θ=1 )
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