Question

Sin ²Q + Cos ²Q = 1 ​

Answer 1

Answer:

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Answer 2

Answer:

$sin {}^{2} q \: + \: cos {}^{2} q = 1$

$sin \: 0 = \frac{9}{c}$

$cos \: 0 = \frac{b}{c}$

$sin {}^{2} + cos {}^{2} 0 = \frac{a {}^{2} }{c ^{2} } + \frac{b {}^{2} }{c {}^{2} } = \frac{a {}^{2} + b {}^{2} }{c {}^{2} }$

$a {}^{2} + b {}^{2} = c {}^{2} so \: \frac{a {}^{2} + b {}^{2} }{c {}^{2} } = 1$

$0 = (0 \times \frac{\pi}{2} )$

$sin(0 + \pi) = - sin(0)$

$cos(0 + \pi) = - cos(0)$

$sin( - 0) = - sin(0)$

$cos( - 0) = cos(0)$

$sin {}^{2} (0 + \pi) \: + cos {}^{2} (0 + \pi) \: = ( - sin0 {}^{2} )$

$(a + b) {}^{2} = c {}^{2} + 4. \frac{1}{2} ab$

$a {}^{2} + 2ab + b {}^{2} = c {}^{2} + 2ab$

$a {}^{2} + b {}^{2} = c {}^{2}$

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