This question will surely come on board examination plz help me to solve it...
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HridayAg0102:
yes
hmmm
that's what we are saying
hmm
I hv tried all methods possible
answer sare methods se ek hi ayega bhai
haa
mera Lhs and Rhs are posted in the answer
plz see there
The person who asked this question also has derived same answers
Answers
Answered by
2
Heya Frnd........,☺
There may be some mistake in the question as the LHS & RHS are not equal.
Here,
LHS = 1 - 6Sin⁴∅Cos⁴∅
& RHS = 1 - 3Sin²∅Cos²∅.
______PLZ CORRECT & REPOST IT ......
There may be some mistake in the question as the LHS & RHS are not equal.
Here,
LHS = 1 - 6Sin⁴∅Cos⁴∅
& RHS = 1 - 3Sin²∅Cos²∅.
______PLZ CORRECT & REPOST IT ......
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Answered by
1
sin ^6theta + cos ^6thta = 1-3(sin theta cos theta )^2
=1-3/4*(2sin theta cos theta )^2
= 4-3(sin 2theta )^2/4.....1
now,
(sin theta + cos theta)^2=x^2
1+sin 2 theta = x^2
sin 2 theta = x^2 -1
put value of sin 2theta in ...1
=4-3(x^2-1)^2/4
this is Ur required result
=1-3/4*(2sin theta cos theta )^2
= 4-3(sin 2theta )^2/4.....1
now,
(sin theta + cos theta)^2=x^2
1+sin 2 theta = x^2
sin 2 theta = x^2 -1
put value of sin 2theta in ...1
=4-3(x^2-1)^2/4
this is Ur required result
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ur welcome nikki57
same from me ☺
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