Question

Please help me to do this sum....
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Answer 1

Answer:

hey here is your answer

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Step-by-step explanation:

$so \: here \\ 4 \: cos {}^{2} x - 1 = 0 \\ ie \: 4 \: cos {}^{2} x = 1 \\ moreover \: here \: \: 0 \leqslant x \leqslant 90$

$so \: thus \: then \\ cos {}^{2} x = 1/4 \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ cos \: x = 1/2 \: \: or \: \:cos \: x = - 1/2$

$but \: as \: here \: x \: > 0 \\ \: cos \: x = - 1/2 \: is \: absurd \\ hence \:cos \: x = 1/2 \\ so \: we \: know \: that \\ cos \: 60 = 1/2 \\ x = 60 \: degrees$

$now \: we \: know \: that \: \\ according \: to \: trigonometric \: identity \\ sin {}^{2} x + cos {}^{2} x = 1 \\ hence \: then \\ sin {}^{2}x + cos {}^{2} x = 1$

$so \: as \: x = 60 \\ we \: know \: that \: \\ sin \: 60 = \sqrt{3} /2 \\ hence \: sin \: x = \sqrt{3} /2$

$now \: thus \: substituting \: values \: of \: sin \: x \: and \: cos \: x \: in \: cos {}^{2} x - sin {}^{2} x \\ we \: get \\ cos {}^{2} x - sin {}^{2} x \\ = (1/2) {}^{2} - ( \sqrt{ 3} /2) {}^{2} \\ = (1/4) - (3/4) \\ 1 - 3/4 \\ = - 2/4 \\ = - 1/2 \\ hence \: \: cos {}^{2} x - sin {}^{2} x = - 1/2$