(sin^(8)75^(@)-cos^(8)75^(@)) = -Turito

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$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

Maths-General

$\frac{7 \sqrt{3}}{8}$
$\frac{7 \sqrt{3}}{16}$
$\frac{3 \sqrt{3}}{16}$
$\frac{3 \sqrt{3}}{8}$

Answer:The correct answer is: $fraction numerator 7 square root of 3 over denominator 16 end fraction$

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$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

$\frac{7 \sqrt{3}}{8}$
$\frac{7 \sqrt{3}}{16}$
$\frac{3 \sqrt{3}}{16}$
$\frac{3 \sqrt{3}}{8}$

Answer:The correct answer is: $fraction numerator 7 square root of 3 over denominator 16 end fraction$

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$open parentheses sin to the power of 8 space 75 to the power of ring operator minus cos to the power of 8 space 75 to the power of ring operator close parentheses$ =

Answer:The correct answer is: $fraction numerator 7 square root of 3 over denominator 16 end fraction$

The minimum value $bold m bold space 2 cos space theta plus fraction numerator 1 over denominator sin invisible function application theta end fraction plus square root of 2 tan space theta space text in end text open parentheses 0 comma pi over 2 close parentheses$ is

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In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

In the figure $A equals 30 to the power of ring operator end exponent & E B equals 50 to the power of ring operator end exponent$ then ACD=

If AB=AC and $A C D equals 115 to the power of ring operator end exponent$ then $stack B with _ below A C equals$

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In the figure $P Q greater than P R comma$ QS and RS are the bisectors of $stack ∣ Q with _ below$ and $stack R with _ below$ respectively then

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$sin space 20 to the power of ring operator open parentheses 4 plus sec space 20 to the power of ring operator close parentheses$ =

$3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator space$ =

$3 tan to the power of 6 space 10 to the power of ring operator minus 27 tan to the power of 4 space 10 to the power of ring operator plus 33 tan squared space 10 to the power of ring operator space$ =

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If $4 n alpha equals pi comma$ then the numerical vlaue of $tan space alpha tan space 2 alpha tan space 3 alpha horizontal ellipsis horizontal ellipsis tan space left parenthesis 2 n minus 1 right parenthesis alpha$ is equal to

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