Question

Prove that :-
${ \cos }^{2} 45 \degree \: - \: \: { \sin }^{2} 15 \degree \: = \: \frac{ \sqrt{3} }{4}$
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Answer 1

Answer:

12−(12–√×3–√2−12–√×12)2

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)= 12−((3–√−1)28)

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)= 12−((3–√−1)28)=12−(3–√−1)28

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)= 12−((3–√−1)28)=12−(3–√−1)28=12−4−23–√8

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)= 12−((3–√−1)28)=12−(3–√−1)28=12−4−23–√8=12−2−3–√4

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)= 12−((3–√−1)28)=12−(3–√−1)28=12−4−23–√8=12−2−3–√4=2−(2+3–√)4

12−(12–√×3–√2−12–√×12)2 =12−(3–√22–√−122–√)2=12−(3–√−122–√)= 12−((3–√−1)28)=12−(3–√−1)28=12−4−23–√8=12−2−3–√4=2−(2+3–√)4=3–√4=RHS

Answer 2

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