CBSE BOARD X, asked by Arny, 1 year ago

This question will surely come on board examination plz help me to solve it...

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HridayAg0102: yes

Anonymous: hmmm

HridayAg0102: that's what we are saying

Anonymous: hmm

HridayAg0102: I hv tried all methods possible

Anonymous: answer sare methods se ek hi ayega bhai

HridayAg0102: haa

HridayAg0102: mera Lhs and Rhs are posted in the answer

HridayAg0102: plz see there

HridayAg0102: The person who asked this question also has derived same answers

Answers

Answered by HridayAg0102

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 ......

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Answered by Anonymous

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

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Anonymous: ur welcome nikki57

HridayAg0102: same from me ☺

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